· contents 0 conventions 3 1 finiteness conditions 5 1.1 finitely generated algebras . . . . . ....

Contents

0 Conventions 3

1 Finiteness conditions 5

1.1 Finitely generated algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Integral extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3 Noetherian rings and modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.4 Application: Invariant rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.5 Graded rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.6 Application: Invariant rings, reprise . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2 Nullstellensatz and Spectrum 19

2.1 Solutions, algebra homomorphisms, and maximal ideals . . . . . . . . . . . . . . . . 20

2.2 A quick word about transcendence bases . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.3 Maximal ideals of finitely generated algebras . . . . . . . . . . . . . . . . . . . . . . 21

2.4 The prime spectrum of a ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3 Exact functors, localization, flatness 29

3.1 Exact, left-exact, right-exact sequences . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.2 Hom and projectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.3 Tensor products and flat modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.4 Localization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4 Decomposition of ideals 41

4.1 Minimal primes and support . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4.2 Associated primes and prime filtrations . . . . . . . . . . . . . . . . . . . . . . . . . 43

4.3 Primary decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

5 Spec and dimension 51

5.1 Dimension and height . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

5.2 Over, up, down theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5.3 Noether normalization and dimension of affine rings . . . . . . . . . . . . . . . . . . 57

6 Dimension, locally 61

6.1 Local rings and NAK . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

6.2 Artinian rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

6.3 Height and number of generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

6.4 The dimension inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

1

2 CONTENTS

7 Hilbert functions 737.1 Hilbert functions of graded rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 737.2 Associated graded rings and general Hilbert functions . . . . . . . . . . . . . . . . . 76

Chapter 0

Conventions

All rings, unless specified otherwise, are commutative and associative with 1 6= 0.All ideals I are assumed to be strict subsets I 6= R. We will call the unit “ideal” an improper

ideal.All modules satisfy the condition 1 ·m = m for all m ∈M .

3

4 CHAPTER 0. CONVENTIONS

Chapter 1

Finiteness conditions

13 Agosto

1.1 Finitely generated algebras

Let ϕ : A→ R be a ring homomorphism. Another piece of terminology for this situation is to saythat R is an A-algebra. An A-algebra is a ring R and a homomorphism ϕ : A→ R; the same ringR with two different maps ϕ,ϕ′ : A→ R yields two different A-algebras.

When R is an A-algebra, we will sometime abuse notation and write a ∈ A for its imageϕ(a) ∈ R; this is not an abuse if A ⊆ R.

A set of elements Λ ⊆ R generates R as an A-algebra if the following equivalent conditionshold: This can be unpackaged more concretely in a number of equivalent ways:

1. The only subring of R containing A and Λ is R itself.

2. Every element of R admits a polynomial expression in Λ with coefficients in ϕ(A).

3. The homomorphism ψ : A[X] → R, where A[X] is a polynomial ring on |Λ| indeterminates,and ψ(xi) = λi, is surjective.

We say that ϕ : A → R is algebra-finite, or R is a finitely generated A-algebra, if there existsa finite set of elements f1, . . . , ft ∈ R that generates R as an A-algebra. The term finite-type isalso used to mean this. A better name might be finitely generatable, since to say that an algebrais finitely generated does not require knowing any actual finite set of generators.

From the discussion above and the first isomorphism theorem, R is a finitely generated A-algebra if and only if R is a quotient of some polynomial ring A[x1, . . . , xd] over A in finitely manyvariables. If R is generated over A by f1, . . . , fd, we will use the notation A[f1, . . . , fd] to denote R.Of course, for this notation to properly specify a ring, we need to understand how these generatorsbehave under the operations. This is no problem if A and f are understood to be contained insome larger ring.

A quick observation: any surjective ϕ is algebra-finite: the target is generated by 1. Sinceany homomorphism ϕ : A → R can be factored as the surjection A → A/ ker(ϕ) followed by theinclusion A/ ker(ϕ) ↪→ R, to understand algebra-finiteness, it suffices to restrict our attention toinjective homomorphisms.

Remark 1.1. Let A ⊆ B ⊆ C be rings. It follows from the definitions that

5

6 CHAPTER 1. FINITENESS CONDITIONS

• A ⊆ B algebra-finite and B ⊆ C algebra-finite =⇒ A ⊆ C algebra-finite, and

• A ⊆ C algebra-finite =⇒ B ⊆ C algebra-finite.

However, A ⊆ C algebra-finite 6=⇒ A ⊆ B algebra-finite.

Example 1.2. Let A = K be a field, and B = K[x, xy, xy2, xy3, · · · ] ⊆ C = K[x, y], where x andy are indeterminates. Any finitely generated subalgebra of B is contained in K[x, xy, . . . , xym] forsome m, since we can write the elements in any finite generating set as polynomial expressions inthe finitely many specified generators of B. But, every element of K[x, xy, . . . , xym] is a K-linearcombination of monomials with the property that the y exponent is no more than m times the xexponent, so this ring does not contain xym+1. Thus, B is not a finitely generated A-algebra.

Let R be an A-algebra and Λ ⊆ R. The ideal of relations on the elements Λ over A is thekernel of the map ψ : A[X] → R, ψ(xi) = λi: this is the polynomial functions with A-coefficientsthat the elements of Λ satisfy. Given an A-algebra R with generators Λ and ideal relations I, wehave R ∼= A[X]/I by the first isomorphism theorem. Thus, if we understand A and generators andrelations for R over A, we can get a pretty concrete understanding of R. If a sequence of elementshas no nonzero relations, we say they are algebraically independent over A.

There are many basic questions about generators that are surprisingly difficult. Let R =C[x1, . . . , xn] and f1, . . . , fn ∈ R. When do f1, . . . , fn generate R? It isn’t too hard to show thatthe Jacobian determinant

det

∂f1∂x1

· · · ∂f1∂xn

.... . .

...∂fn∂x1

· · · ∂fn∂xn

must be a nonzero constant. It is a big open question whether this is in fact a sufficient condition!

1.2 Integral extensions

We will also find it quite useful to consider a stronger finiteness property for maps. Recall that ifϕ : A→ R is a ring homomorphism, then R acquires an A-module structure via ϕ by a · r = ϕ(a)r;this is a particular case of restriction of scalars. We may write ϕR for this A-module if we thinkwe will have trouble remembering the map. Of course, if ϕ is injective and we identify it with theinclusion A ⊆ R, then this A-action is just a · r = ar.

Recall that an A-module M is generated by a set of elements Γ ⊆M if the only submodule ofM that contains Γ is M itself. This also has some equivalent realizations:

1. Γ generates M as an A-module.

2. Every element of M admits a linear combination expression in Γ with coefficients in A.

3. The homomorphism θ : A⊕Y →M , where A⊕Y is a free A-module on |Γ| basis elements, andθ(yi) = γi, is surjective.

We recall that a basis for an R-module is a generating set Γ such that∑

i aiγi = 0 implies allai = 0, for ai ∈ R, γi ∈ Γ; a free module is a module that admits a basis. Every ring that is not afield admits a non-free module, namely the cyclic module R/I for an ideal I ⊆ R.

We use the notation M =∑

γ∈ΓAγ to indicate that M is generated by Γ as a module. Wesay that ϕ : A → R is module-finite if R is a finitely-generated A-module. This is also called

1.2. INTEGRAL EXTENSIONS 7

simply finite in the literature; we will see why in a few weeks, but we’ll stick with the unambiguous“module-finite.”

As with algebra-finiteness, surjective maps are always module-finite in a trivial way. Thus, itsuffices to understand this notion for ring inclusions.

The notion of module-finite is much stronger than algebra-finite, since a linear combination isa very special type of polynomial expression.

Example 1.3. 1. If K ⊆ L are fields, L is module-finite over K just means that L is a finitefield extension of K.

2. The Gaussian integers Z[i] satisfy the well-known property (or definition, depending on yoursource) that any element z ∈ Z[i] admits a unique expression z = a+ bi with a, b ∈ Z. Thatis, Z[i] is generated as a Z-module by {1, i}; moreover, they form a free module basis!

3. If R is a ring and x an indeterminate, R ⊆ R[x] is not module-finite. Indeed, R[x] is a freeR-module on the basis {1, x, x2, x3, . . . }.

4. Another map that is not module-finite is the inclusion of K[x] ⊆ K[x, 1/x]. Note that anyelement of K[x, 1/x] can be written in the form f(x)/xn for some f and n. Then, any

finitely generated K[x]-submodule M of K[x, 1/x] is of the form M =∑

ifi(x)xni ·K[x]; taking

N = max{ni | i}, we find that M ⊆ 1/xN ·K[x] 6= K[x, 1/x].

Lemma 1.4. If R ⊆ S is module-finite, and N is a finitely generated S-module, then N is a finitelygenerated R-module by restriction of scalars. In particular, the composition of two module-finitering maps is module-finite.

Proof. Let S =∑r

i=1Rai and N =∑s

j=1 Sbj . Then, N =∑r

i=1

∑sj=1Raibj : given n =

∑sj=1 sjbj ,

rewrite each sj =∑r

i=1 rijai and substitute to get t =∑r

i=1

∑sj=1 rijaibj as an R-linear combination

of the aibj .

In field theory, there is a close relationship between (vector space-)finite field extensions andalgebraic equations. The situation for rings is similar.

Definition 1.5 (Integral element/extension). Let ϕ : A→ R be a ring homomorphism, and r ∈ R.The element r is integral if there are elements a0, . . . , an−1 ∈ A such that

rn + an−1rn−1 + · · ·+ a1r + a0 = 0;

i.e., r satisfies a equation of integral dependence over A.

We say that R is integral over A if every r ∈ R is integral over A.

Like our other smallness hypotheses for maps, we see that r ∈ R is integral over A if and only ifr is integral over the subring ϕ(A) ⊆ R. Again, we can restrict our focus to inclusion maps A ⊆ R.

Evidently, integral implies algebraically dependent, and the condition that there exists an equa-tion of algebraic dependence that is monic is stronger in the setting of rings.

Proposition 1.6. Let A ⊆ R be rings.

1. If r ∈ R is integral over A then A[r] is module-finite over A.

2. If r1, . . . , rt ∈ R are integral over A then A[r1, . . . , rt] is module-finite over A.


Proof. 1. Suppose r is integral over A, satisfying the equation rn+an−1rn−1 + · · ·+a1r+a0 = 0.

Then A[r] =∑n−1

i=0 Ari. Indeed, given a polynomial in p(r) of degree ≥ n, we can use the

equation above to rewrite the leading term amrm as −amrm−n(an−1rn−1 + · · · + a1r + a0),

and decrease the degree in r.

2. Write A0 := A ⊆ A1 := A[r1] ⊆ A2 := A[r1, r2] ⊆ · · · ⊆ At := A[r1, . . . , rt]. Note thatri is integral over Ai−1: use the same monic equation of ri over A. Then, the inclusionA ⊆ A[r1, . . . , rt] is a composition of module-finite maps, hence is module-finite.

The name “ring” is roughly based on this idea: in an extension as above, the powers wraparound (like a ring).

15 Agosto

We will need a linear algebra fact. The classical adjoint of an n × n matrix B = [bij ] is the

matrix adj(B) with entries adj(B)ij = (−1)i+j det(Bji), where Bji is the matrix obtained from Bby deleting its jth row and ith column. You may remember this matrix from Cramer’s rule.

Lemma 1.7 (Determinant trick). Let R be a ring, B ∈Mn×n(R), v ∈ R⊕n, and r ∈ R.

1. adj(B)B = det(B)In×n.

2. If Bv = rv, then det(rIn×n −B)v = 0.

Proof. 1. When R is a field, this is a basic fact. We deduce the case of a general commutativering from the field case.

The ring R is a Z-algebra (every ring is a Z-algebra, but generally not finitely generated assuch), so we can write R as a quotient of some polynomial ring Z[X]. Let ψ : Z[X] → Rbe a surjection, let aij ∈ Z[X] be such that ψ(aij) = bij , and let A = [aij ]. Note thatψ(adj(A)ij) = adj(B)ij and ψ((adj(A)A)ij) = (adj(B)B)ij , since ψ is a homomorphism,and the entries are the same polynomial functions of the entries of the matrices A and B,respectively. Thus, it suffices to establish the lemma in the case R = Z[X]. Now, R = Z[X]is an integral domain, hence a subring of its fraction field. Since both sides of the equationlive in R and are equal in the fraction field (by linear algebra) they are equal in R.

2. We have (rIn×n −B)v = 0, so det(rIn×n −B)v = adj(rIn×n −B)(rIn×n −B)v = 0.

Theorem 1.8 (Module finite implies integral). Let A ⊆ R be module-finite. Then R is integralover A.

Proof. Let r ∈ R. The idea is to show that multiplication by r, realized as a linear transformationover A, satisfies the characteristic polynomial of that linear transformation.

Write R =∑t

i=1Ari. We may assume that r1 = 1, since we can add module generators. Byassumption, we can find aij ∈ A such that

rri =

t∑j=1

aijrj

for each i. Let C = [aij ], and v be the column vector (r1, . . . , rt). We then have rv = Cv, so bythe determinant trick, det(rIn×n − C)v = 0. In particular, det(rIn×n − C) = 0. Exapanding as apolynomial in r, this is a monic equation with coefficients in A.

1.3. NOETHERIAN RINGS AND MODULES 9

Corollary 1.9 (Characterization of module-finite extensions). Let A ⊆ R be rings. R is module-finite over A if and only if R is integral and algebra-finite over A.

Proof. (⇒): A generating set for R as an A-module serves as a generating set as an A-algebra.The rest of this direction comes from the previous theorem. (⇐): If R = A[r1, . . . , rt] is integralover A, so that each ri is integral over A, then R is module-finite over A by Proposition 1.6.

Corollary 1.10. If R is generated over A by integral elements, then R is integral. Thus, if A ⊆ S,the set of elements of S that are integral over A form a subring of S.

Proof. Let R = A[Λ], with λ integral over A for all λ ∈ Λ. Given r ∈ R, there is a finite subsetL ⊆ Λ such that r ∈ A[L]. By the theorem, A[L] is module-finite over A, and r ∈ A[L] is integralover A.

For the latter statement,

{integral elements} ⊆ A[{integral elements}] ⊆ {integral elements},

so equality holds throughout, and {integral elements} is a ring.

Definition 1.11. If A ⊆ R, the integral closure of A in R is the set of elements of R that areintegral over A.

Example 1.12. 1. Let R = C[x, y] ⊆ S = C[x, y, z]/(x2 + y2 + z2). The ring S is module-finiteover R: indeed, S is generated over R as an algebra by one element z that is integral over R.

2. Let R = C[u, v] ⊆ S = C[u, v, w]/(u2 +vw). (Note that this S is isomorphic to the previous Sby the map u 7→ x, v 7→ y+ iz, w 7→ y− iz, and this R corresponds to the polynomial subringC[x, y + iz].) We claim that S is not integral and hence not module-finite over R. Indeed,the minimal polynomial of w over the fraction field of R is f(t) = vt+u2. Any equation thatw satisfies is a C(u, v)[t]-multiple of this: write g(t) = f(t)h(t) with g(t) ∈ C(u, v)[t] monic.By Gauss’ lemma, there is some a ∈ C(u, v) such that a−1f(t), ah(t) ∈ C[u, v][t]. Since theleading coefficient of h is v−1, the numerator of a must be a multiple of v when written inlowest terms. But this contradicts that a−1f(t) ∈ C[u, v][t].

3. Not all integral extensions are module-finite. Let K = K, and consider the ring R =K[x, x1/2, x1/3, x1/4, x1/5, . . . ] ⊆ K(x). Clearly R is generated by integral elements over K[x],but is not algebra-finite over K[x]. (Prove it!)

Remark 1.13. Let A ⊆ B ⊆ C be rings. As with algebra-finiteness, it follows from the definitionsthat

• A ⊆ B module-finite and B ⊆ C module-finite =⇒ A ⊆ C module-finite, and

• A ⊆ C module-finite =⇒ B ⊆ C module-finite,

but again, A ⊆ C module-finite 6=⇒ A ⊆ B module-finite.

1.3 Noetherian rings and modules

We now discuss an absolute, rather than relative, finiteness condition for a ring R.

Definition 1.14 (Noetherian ring). A ring R is Noetherian if every ascending chain of idealsI1 ⊆ I2 ⊆ I3 ⊆ · · · eventually stabilizes: there is some N for which In = In+1 for all n > N .


This condition also admits some equivalences.

Proposition 1.15 (Equivalences for Noetherian ring). The following are equivalent for a ring R.

1. R is a Noetherian ring.

2. Every nonempty family of ideals has a maximal element (under containment).

3. Every ascending chain of finitely generated ideals of R eventually stabilizes.

4. Given any generating set S for an ideal I, the ideal I is generated by a finite subset of S.

5. Every ideal of R is finitely generated.

Proof. (1)⇒(2): We prove the contrapositive. Suppose there is a nonempty family of ideals withno maximal element. This means that we can inductively keep choosing larger ideals from thisfamily to obtain an infinite properly ascending chain.

(2)⇒(1): Clear.

(1)⇒(3): Clear.

(3)⇒(4): We prove the contrapositive. Suppose that there is an ideal I and generating set Ssuch that no finite subset of S generates I. For any finite S′ ⊆ S we have (S′) ( (S) = I, so thereis some s ∈ Sr (S′). Thus, (S′) ( (S′∪{s}). Inductively, we can continue this to obtain an infiniteproper chain of finitely generated ideals, contradicting (3).

(4)⇒(5): Clear.

(5)⇒(1): Given an ascending chain of ideals I1 ⊆ I2 ⊆ I3 ⊆ · · · let I =⋃n∈N In. The ideal I

is finitely generated, say I = (a1, . . . , at), and since each ai is in some Ini , there is an N such thateach ai is in IN . But then In = I = IN for all n > N .

Example 1.16. 1. If K is a field, the only ideals in K are (0) and (1) = K, so K is Noetherian.

2. If R is a PID, then R is Noetherian. Every ideal is finitely generated!

3. As a special case of the previous, consider the ring of germs of complex analytic functionsnear 0,

C{z} := {f(z) ∈ CJzK | f is analytic on a neighborhood of z = 0}.

This ring is a PID: every ideal is of the form (zn), since any f ∈ C{z} can be written aszng(z) with g(z) 6= 0, and any such g(z) is a unit in C{z}.

4. A ring that is not Noetherian is a polynomial ring in infinitely many variables over a field K:the ascending chain of ideals (x1) ⊆ (x1, x2) ⊆ (x1, x2, x3) ⊆ · · · does not stabilize.

5. Another ring that is not Noetherian is the ring R = K[x, x1/2, x1/3, x1/4, x1/5, . . . ] ⊆ K(x)from earlier. A nice ascending chain of ideals is

(x) ⊆ (x1/2) ⊆ (x1/3) ⊆ (x1/4) ⊆ · · · .

6. A variation on the last example: the ring of nonnegatively valued Puiseux series: R =⋃n∈NCJz1/nK ⊆ C((z)).1

1In fact, the algebraic closure of the field of Laurent series C((z)) is⋃n∈N C((z1/n)) = R[1/t].

1.3. NOETHERIAN RINGS AND MODULES 11

7. The ring of continuous real-valued functions C(R,R) is not Noetherian: the chain of idealsIn = {f(x) | f |[−1/n,1/n] ≡ 0} is increasing and proper. The same construction shows thatthe ring of infinitely differentiable real functions C∞(R,R) is not Noetherian: properness ofthe chain follows from, e.g., Urysohn’s lemma (though it’s not too hard to find functionsdistinguishing the ideals in the chain). Note that if we asked for analytic functions instead ofinfinitely-differentiable functions, every element of the chain would be the zero ideal!

Definition 1.17 (Noetherian module). An R-module M is Noetherian if every ascending chain ofsubmodules of M eventually stabilizes.

There are analogous criteria for modules to (1)–(5) above namely:

Proposition 1.18 (Equivalences for Noetherian module). The following are equivalent for a moduleM :

1. M is a Noetherian module.

2. Every nonempty family of submodules has a maximal element.

3. Every ascending chain of finitely generated submodules of M eventually stabilizes.

4. Given any generating set S for a submodule N , the submodule N is generated by a finitesubset of S.

5. Every submodule of M is finitely generated.

In particular, a Noetherian module must be finitely generated.

We observe that if R is Noetherian, and I is an ideal of R, then R/I is Noetherian as well, sincethere is an order-preserving bijection

{ideals of R that contain I} ↔ {ideals of R/I}.

Lemma 1.19. Let M be a module, and M ′,M ′′, N be submodules of M .

1. Then M ′ = M ′′ if and only if M ′ ∩N = M ′′ ∩N and M ′/(M ′ ∩N) = M ′′/(M ′′ ∩N).

2. M is Noetherian if and only if N and M/N Noetherian.

Proof. 1. (⇒) is clear. Suppose that M ′ ( M ′′ and M ′ ∩ N = M ′′ ∩ N . Then, there is somem ∈M ′′ rM ′. We claim that the class of m in M ′′/(M ′′ ∩N) is not equal to the class of m′

in M ′′/(M ′′ ∩N) for any m′ ∈M ′. Indeed, if it were, then m−m′ ∈M ′′ ∩N = M ′ ∩N , sothat m ∈M ′ +M ′ ∩N = M ′, a contradiction.

2. A chain of submodules of N is a chain of submodules of M , and a (proper) chain of submodulesof M/N lifts to a (proper) chain of submodules of M , so M Noetherian implies N and M/Nare. Conversely, if N and M/N are Noetherian, then for any chain

M0 ⊆M1 ⊆M2 ⊆ · · ·

of submodules of M , the chains

(M0 ∩N) ⊆ (M1 ∩N) ⊆ (M2 ∩N) ⊆ · · ·

in N andM0

M0 ∩N⊆ M1

M1 ∩N⊆ M2

M2 ∩N⊆ · · ·

in M/N stabilize eventually. If both chains are stable past index n, then the chain in M isstable past index n by part 1.


Proposition 1.20. Let R be a Noetherian ring. Then M is a Noetherian module if and only if Mis finitely generated.

Consequently, if R is Noetherian, then any submodule of a finitely generated R-module is alsofinitely generated.

Proof. If M is Noetherian, it (and all of its submodules) is finitely generated by the equivalencesabove.

Now let R be Noetherian and M be f.g.. First, we show that the free module R⊕n =⊕n

i=1Rei isNoetherian for all n ∈ N by induction on n. For the base case, we note that R is a Noetherian ringiff the free cyclic module R⊕1 is a Noetherian module, since ideals of R correspond to submodulesof R⊕1. The inductive step follows since we have an isomorphism (

⊕ni=1Rei)/Ren

∼=⊕n−1

i=1 Rei.Now, a finitely generated module M is quotient of a finitely generated free module, so is Noetherianby the previous lemma.

20 Agosto

It is now clear that a module-finite extension R of a Noetherian ring A is Noetherian: R is aNoetherian A-module, and any ideal of R is an A-submodule of R, so an ascending chain necessarilystabilizes.

Here is a stronger statement:

Theorem 1.21 (Hilbert Basis Theorem). Let A be a Noetherian ring. Then A[x1, . . . , xd] andAJx1, . . . , xdK are Noetherian.

Proof. We give the proof for polynomial rings, and indicate the difference in the power seriesargument.

By induction on d, we reduce to the case d = 1. Let I ⊆ A[x], and let

J = {a ∈ A | ∃ axn + lower order terms (wrt x) ∈ I}.

This is easily seen to be an ideal of A, which is finitely generated by hypothesis; let J = (a1, . . . , at).Pick f1, . . . , ft ∈ A[x] such that the leading coefficient of fi is ai, and set i′ = (f1, . . . , ft). LetN = maxi deg fi.

Given f ∈ I of degree greater than N , we can cancel off the leading term of f by subtractinga suitable multiple of some fi, so any f ∈ I can be written as g + h with g ∈ I ∩

∑Ni=0Ax

i and

h ∈ I ′. Since I ∩∑N

i=0Axi is a submodule of a finitely generated free A-module, it is also finitely

generated as an A-module. Given such a generating set, we can clearly write any such f as anA[x]-linear combination of these generators and the fi’s.

In the power series case, take J to be the coefficients of lowest degree terms.

Corollary 1.22. If A is a Noetherian ring, then any finitely generated A-algebra is Noetherian.In particular, any finitely generated algebra over a field is Noetherian.

The converse to this statement is false: there are lots of Noetherian rings that are not f.g.algebras over a field. For example, C{z} is not algebra-finite over C. You will show (modulosome steps) that rings of power series and rings of complex functions analytic near a point are bothNoetherian. We will also see huge classes of easy examples once we switch to localization. However,we’ll see a converse in a very special case soon.

Now, we prove a technical theorem that relates all of our finiteness notions. The statement isa bit complicated, but the result will be pretty useful.

1.4. APPLICATION: INVARIANT RINGS 13

Theorem 1.23 (Artin-Tate Lemma). Let A ⊆ B ⊆ C be rings. Assume that

• A is Noetherian,

• C is module-finite over B, and

• C is algebra-finite over A.

Then, B is algebra-finite over A.

Proof. Let C = A[f1, . . . , fr] and C =∑s

i=1Bgi. Then,

fi =∑

bijgj and gigj =∑

bijkgk

for some elements bij , bijk ∈ B. Let B0 = A[{bij , bijk}] ⊆ B. Since A is Noetherian, so is B0.We claim that C =

∑si=1B0 gi. Given an element c ∈ C, write c as a polynomial expression in

f . We have that c ∈ A[{bij}][g1, . . . , gs]. Then, using the equations for gigj , we can write c in theform required.

Now, since B0 is Noetherian, C is a finitely generated B0-module, and B ⊆ C, then B isa finitely generated B0-module, too. In particular, B0 ⊆ B is algebra-finite. We conclude thatA ⊆ B is algebra-finite, as required.

1.4 Application: Invariant rings

Question 1.24. Given a (finite) set of symmetries, one can consider the collection of polynomialfunctions that are fixed by all of them. Is there a finite set of fixed polynomials such that any fixedpolynomial can be expressed in terms of them?

Let G be a group acting on a ring R, or just as well, a group of automorphisms of R. The maincase we have in mind is when R = K[x1, . . . , xd] is a polynomial ring over a field. We are interestedin the set of elements that are invariant under the action

RG := {r ∈ R | g(r) = r for all g ∈ G}.

If r, s ∈ RG, then

r + s = g(r) + g(s) = g(r + s) and rs = g(r)g(s) = g(rs) for all g ∈ G,

since each g is a homomorphism. Thus, RG is a subring of R.We note that if G = 〈g1, . . . , gt〉, then r ∈ RG if and only if gi(r) = r for i = 1, . . . , t.We will especially be generated by linear actions: G acts linearly on a polynomial ring R =

K[x1, . . . , xn] if every element of g satisfies g · k = k for all k ∈ K and g · xi is a linear form in Rfor all i.

Example 1.25 (Standard representation of the symmetric group). Let Sd be the symmetric groupon d letters acting on R = K[x1, . . . , xd] via σ(xi) = xσi .

For example, if d = 3, then f = x21 + x2

2 + x23 is invariant, while g = x2

1 + x1x2 + x22 + x2

3 is not,since swapping 1 with 3 gives a different polynomial.

It is a theorem that you may have seen this in an earlier algebra class that every elementof RG can be written as polynomial expression in the elementary symmetric polynomials ei =∑I⊆[d], |I|=i

(∏j∈I xj). E.g, f above is e2

1 − 2e2.


Example 1.26 (Roots of unity). Let G = {e, g} act on R = K[x1, . . . , xd] by negating the variables:g · xi = −xi for all i, so g · f(x) = f(−x). Suppose that the characteristic of K is not 2, so −1 6= 1.Given a general f , we can write it as a sum of its homogeneous pieces: that is,

f = fr + fr−1 + · · ·+ f1 + f0,

where each fi is a sum of monomials of degree i. We have g(fi) = (−1)ifi, so

g(f) = (−1)rfr + (−1)r−1fr−1 + · · · − f1 + f0,

which differs from f unless every homogeneous piece of f has even degree. That is,

RG = {f ∈ R | every term of f has even degree}.

This computation readily generalizes to the case of a field K that contains t-th roots of unity,and a cyclic group G = 〈g〉 of order t acting on R by the rule g(xi) = ζt · xi for all i. In this case,we have

RG = {f ∈ R | every term of f has degree a multiple of t}.This ring is called the t-th Veronese ring . As an algebra, this ring is generated by the monomialsof degree t.

Example 1.27. Some of the most interesting examples come from infinite groups G. Let X =X2×3, be a matrix of indeterminates, and C[X] be the polynomial ring on those indeterminates.The group SL2(C) acts on the matrix X by left multiplication. Write Xi for the matrix X withcolumn i removed. If g ∈ SL2(C), then gXi = (gX )i, since left multiplication by g can be computedcolumn-by-column. Thus,

det((gX )i) = det(gXi) = det(g) det(Xi) = det(Xi).

We find that the 2 × 2 minors of X (x11x22 − x12x21, x11x23 − x13x21, x12x23 − x13x22) must beinvariant functions. This readily generalizes to m × n matrices with m ≤ n and maximal minors.It is harder to show that these maximal minors generate all of the invariants.

21 Agosto

We now want to move towards answering our question. We will use our various notions offiniteness.

Proposition 1.28. Let K be a field, R be a finitely-generated K-algebra, and G a finite group ofautomorphisms of R that fix K. Then, RG ⊆ R is module-finite.

Proof. We will show that R is algebra-finite and integral over RG.First, since R is generated by a finite set as a K-algebra, and K ⊆ RG, it is generated by the

same finite set as an RG-algebra as well.Now, for r ∈ R, consider the polynomial Fr(t) =

∏g∈G(t−g(r)) ∈ R[t]. Clearly g(Fr(t)) = Fr(t),

where G fixes t. Thus, Fr(t) ∈ RG[t]. The leading term (with respect to t) is t|G|, so Fr(t) is monic.Thus, r is integral over RG. Therefore, R is integral over RG.

Theorem 1.29 (Noether’s finiteness theorem for invariants of finite groups). Let K be a field, Rbe a polynomial ring over K, and G be a finite group acting K-lienarly on R. Then RG is a finitelygenerated K-algebra.

Proof. Observe that K ⊆ RG ⊆ R, that K is Noetherian, K ⊆ R is algebra-finite, and RG ⊆ R ismodule-finite. Thus, by the Artin-Tate Lemma, we are done!

1.5. GRADED RINGS 15

1.5 Graded rings

A useful bit of extra structure that one commonly encounters, and that we have already used, even,is that of a grading on a ring.

Definition 1.30. Let R be a ring, and T be a monoid. The ring R is T -graded if there exists adirect sum decomposition of R as an abelian group indexed by T : R =

⊕a∈M Ra such that, for any

a, b ∈ T , and any r ∈ Ra, s ∈ Rb, one has rs ∈ Ra+b.

An element that lies in one of the summands Ra is said to be homogeneous of degree a; weoften use |r| to denote the degree of a homogeneous element r.

Definition 1.31. Let R and S be T -graded rings (same grading monoid). A ring homomorphismϕ : R→ S is degree-preserving if ϕ(Ra) ⊆ Sa for all a ∈ T .

By definition, an element in a graded ring is, in a unique way, a sum of homogeneous elements,which we call its homogeneous components or graded components.

Example 1.32. 1. If K is a field, and R = K[x1, . . . , xn] is a polynomial ring, then there is anN-grading on R where Rd is the K-vector space with basis given by monomials xα1

1 · · ·xαnnwith

∑i αi = d. Of course, this is the notion of degree familiar from grade school. This is

called the standard grading .

2. With K and R as above, for any (β1, . . . , βn) ∈ Nn, one can give a different N-grading on R byletting xi have degree βi for some integers βi; we call this a grading with weights (β1, . . . , βn).

For example, in K[x1, x2], x21 + x3

2 is not homogeneous in the standard grading, but is homo-geneous of degree 6 under the N-grading with weights (3, 2).

3. Again with K and R as above, R admits an Nn-grading, with R(d1,...,dn) = K · xd11 · · ·xdnn .This is called the fine grading .

4. Let Γ ⊆ Nn be a subsemigroup. Then⊕

γ∈ΓKxγ ⊆ K[x] is an Nn-graded subring. Conversely,

every Nn-graded subring of K[x1, . . . , xn] is of this form. (Check it!)

5. If R is a graded ring, and G is a group acting on R by degree-preserving automorphisms, thenRG is a graded subring of R. (I.e., RG is graded with respect to the same grading monoid.) Inparticular, if G acts K-linearly on a polynomial ring over K, the invariant ring is N-graded.

Definition 1.33. An ideal I in a graded ring R is called homogeneous if it is generated by homo-geneous elements.

We observe that an ideal is homogeneous if and only if, for any f ∈ R, one has f ∈ I if andonly if every homogeneous component of f lies in I. To see “if,” take a generating set {fλ}Λ for I;all of its homogeneous components of each fλ lie in I, and each fλ lies in the ideal generated bythese components. Thus the set of components generates I. The other direction is also easy.

We now observe the following:

Lemma 1.34. Let R be an T -graded ring, and I be a homogeneous ideal. Then R/I is alsoT -graded.

Example 1.35. 1. The ring R = K[w, x, y, z]/(w2x + wyz + z3, x2 + 3xy + 5xz + 7yz + 11z2)admits an N-grading with |w| = |x| = |y| = |z| = 1.


2. The ring R = K[x, y, z]/(x2 + y3 + z5) does not admit a grading with |x| = |y| = |z| = 1, butdoes admit a grading with |x| = 15, |y| = 10, |z| = 6.

Definition 1.36. Let R be a T -graded ring, and M a module. The module R is T -graded if thereexists a direct sum decomposition of M as an abelian group indexed by T : M =

⊕a∈T Ma such

that, for any a, b ∈ T , and any r ∈ Ra,m ∈ Rb, one has rm ∈Ma+b.

The notions of homogeneous element of a module and degree of a homogeneous element of amodule take the obvious meanings. A notable abuse of notation: we will often talk about Z-gradedmodules over N-graded rings, and likewise.

We observed earlier an important relationship between algebra-finiteness and Noetherianitythat followed from the Hilbert basis theorem: if R is Noetherian, then any algebra-finite extensionof R is also Noetherian. There isn’t a converse to this in general: there are lots of algebras overfields K that are Noetherian but not algebra-finite over K. However, for graded rings, this converserelation holds.

Proposition 1.37. Let R be an N-graded ring, and f1, . . . , fn be homogeneous elements of positivedegree. Then f1, . . . , fn generate the ideal R+ :=

⊕d>0Rd if and only if f1, . . . , fn generate R as

an R0-algebra.Consequently, R is Noetherian if and only if R0 is Noetherian and R is algebra-finite over R0.

Proof. If R = R0[f1, . . . , fn], then any element r ∈ R+ can be written as a polynomial expressionr = P (f1, . . . , fn) with P ∈ R0[x] and P with no constant term. Each monomial of P then is amultiple of some xi, so that r ∈ (f1, . . . , fn).

To show that R+ = (f1, . . . , fn) implies R = R0[f1, . . . , fn], it suffices to show that any ho-mogeneous element r ∈ R can be written as a polynomial expression in the f ’s with coefficientsin R0. We induce on the degree of r, with degree 0 as a trivial base case. For r homogeneous ofpositive degree, r ∈ R+ so we can write r = a1f1 + · · ·+ anfn; moreover, we can do so with each aihomogeneous of degree |r| − |fi|. By the induction hypothesis, each ai is a polynomial expressionin the f ’s, so we are done.

For the consequently statement, if R0 is Noetherian and R algebra-finite over R0, then R isNoetherian by Hilbert Basis. If R is Noetherian, R0 is Noetherian, since it is isomorphic to R/R+,and R is algebra-finite over R0 since R+ is generated as an ideal by finitely many homogeneouselements by Noetherianity, so by the first statement, we get a finite algebra generating set for Rover R0.

1.6 Application: Invariant rings, reprise

With the basics of graded rings in hand, we can give a different proof of the finite generation ofinvariant rings that works under different hypotheses. The proof we will discuss now is essentiallyHilbert’s proof. We need another notion that is very useful in commutative algebra.

Definition 1.38. Let R be an A-algebra, where φ : A→ R is the map from A to R. We say thatA is a direct summand of R if there is an A-module homomorphism π : R→ A such that π ◦ φ isthe identity on A.

First, observe that the condition implies that φ must be injective, so we can assume that A ⊆ R(perhaps after renaming elements). Then the condition on π is that π(ar) = aπ(r) for all a ∈ Aand r ∈ R and that π|A is the identity. We call the map π the splitting of the inclusion. Note thatif π : R→ A is A-linear, if π(1) = 1, then π is a splitting, since π(a) = π(a · 1) = aπ(1) = a for alla ∈ A.

1.6. APPLICATION: INVARIANT RINGS, REPRISE 17

Lemma 1.39. Let A be a direct summand of R. Then, for any ideal I ⊆ A, we have IR ∩A = I.

Proof. Let π be the splitting of the inclusion. If a ∈ IR ∩ A, we have a =∑t

i=1 rifi, for fi ∈ I.Applying π, we have

a = π(a) = π(

t∑i=1

rifi) =

t∑i=1

π(rifi) =

t∑i=1

π(ri)fi ∈ I.

Proposition 1.40. Let A be a direct summand of R. If R is Noetherian, then so is A.

Proof. Let I1 ⊆ I2 ⊆ I3 ⊆ · · · be a chain of ideals inA. The chain I1R ⊆ I2R ⊆ I3R ⊆ · · · stabilizes,so ∃J,N such that InR = J for n ≥ N . Contracting to A, we get that In = InR ∩ A = J ∩ A forn ≥ N , so the chain stabilizes.

Proposition 1.41. Let K be a field, and R be a polynomial ring over K. Let G be a finite groupacting K-linearly on R. Assume that |G| does not divide the characteristic of K; this condition istrivially satisfied if K has characteristic zero. Then RG is a direct summand of R.

Proof. We consider the map ρ : R→ RG given by ρ(r) = 1|G|∑

g∈G g · r. First, note that the image

of this maps lies in RG, since acting by g permutes the elements in the sum. Now, we claim thatthis is a splitting of the inclusion: let s ∈ RG and r ∈ R. We have

ρ(sr) =1

|G|∑g∈G

g · (sr) =1

|G|∑g∈G

(g · s)(g · r) =1

|G|∑g∈G

s(g · r) = s1

|G|∑g∈G

(g · r) = sρ(r),

so ρ is RG-linear, and for s ∈ RG, ρ(s) = 1|G|∑

g∈G g · s = s.

Theorem 1.42 (Hilbert’s finiteness theorem for invariants). Let K be a field, and R be a polynomialring over K. Let G be a group acting K-linearly on R. Assume that G is finite and |G| does notdivide the characteristic of K (or more generally, that RG is a direct summand of R). Then RG isa finitely generated K-algebra.

Proof. Since G acts linearly, RG is an N-graded subring of R with R0 = K. Since RG is adirect summand of R, RG is Noetherian by the previous proposition. By our characterization ofNoetherian graded rings, RG is finitely generated over R0 = K.

One important thing about this proof is that it applies to many infinite groups. In particular,for any linearly reductive group, including GLn(C),SLn(C), (C×)n, etc., one can construct a maplike ρ that is a splitting.

Chapter 2

Nullstellensatz and Spectrum

27 Agosto

A motivating question to lead us to our next big theorem is the following:

Question 2.1. To what extent is a system of polynomial equations determined by its solution set?

Example 2.2. Let’s consider one polynomial equation in one variable. Over R,Q, or other fieldsthat aren’t algebraically closed, there are many polynomials with an empty solution set. On theother hand, over C, or any algebraically closed field, if f(z) = 0 has solutions z1, . . . , zd, we knowthat f(z) = α(z − z1)a1 · · · (z − zd)ad , so that f is determined up to scalar multiple and repeatedfactors. Given any system of polynomial equations f1 = · · · = ft = 0, we know that z = a is asolution if and only if it is a solution of f = 0, where f is the GCD of f1, . . . , ft.

Now, let A be any ring. Let x = {xγ | γ ∈ Γ} be a set of variables, and F = {fλ | λ ∈ Λ} ⊆ A[x]be a set of polynomials. If R is any A-algebra, we can evaluate any polynomial by plugging inelements of R: given r = {rγ | γ ∈ Γ}, we can evaluate f(r) in R, and determine whether f(r) = 0.In this situation, we define the solution set to be

ZR(F ) := {r ∈ R⊕Γ | f(r) = 0 for all f ∈ F}.

In particular, if we are considering a system of polynomial equations in finitely many variablesx1, . . . , xn, then ZR(F ) ⊆ R⊕n. Observe that if F ⊆ F ′, then ZR(F ′) ⊇ ZR(F ). Also, if r ∈ZR(f1, f2), and g ∈ A[x], then

(f1 + f2)(r) = f1(r) + f2(r) = 0 and (f1g)(r) = f1(r)g(r) = 0,

so ZR(F ) ⊆ ZR( (F ) ), and in tandem with the above, equality holds: ZR(F ) = ZR( (F ) ). In par-ticular, by the Hilbert basis theorem, any system of polynomial equations over a field is equivalentto a system of finitely many polynomial equations!

From now on, we will talk about the solution set of an ideal, rather than of an arbitrary set.Observe that if I = R is an improper ideal, then ZR(I) = ∅, since 1 6= 0.

Example 2.3. Let R = K

[x1 x2 x3

y1 y2 y3

], and I = (∆1,∆2,∆3), the 2 × 2-minors of the matrix.

Thinking of these generators as equations, a solution is just a matrix of rank at most one, so ZK(I)is the set of rank at most one matrices. Let J = (x1, x2, x3), and observe that I ⊆ J . We thenhave that ZK(J) ⊆ ZK(I): this just translates to the fact that a 2× 3 matrix with a zero row hasrank at most 1.

19

20 CHAPTER 2. NULLSTELLENSATZ AND SPECTRUM

2.1 Solutions, algebra homomorphisms, and maximal ideals

We now wish to consider solutions of polynomial equations in great generality. A homomorphismof R-algebras from S to T is a ring homomorphism such that

S // T

R

??__

commutes.

Let R be a ring, and S an R-algebra. Let I ⊆ R[x] be an ideal. Consider the R-algebra R[x]/I.If a ∈ ZS(I) is a solution, so that all f(a) = 0 for all f ∈ I, then the R-algebra homomorphismfrom R[x] → S given by xi 7→ ai descends to an R-algebra map R[x]/I → S. Conversely, anyR-algebra map from R[x]/I to S is determined by the images of a, which must satisfy all of theequations fλ(x) = 0. We summarize:

Proposition 2.4. Let R be a ring, and S an R-algebra. Let I ⊆ R[x] be an ideal. There is abijection

ZS(I) ↔ HomR−alg (R[x]/I, S)

(a1, . . . , ad) 7→ ϕ|R = id, ϕ(xi) = ai.

Now, we focus on polynomial equations over fields.

Proposition 2.5. Let K be a field, and R = K[x1, . . . , xd] be a polynomial ring. There is abijection

Kd ↔ {maximal ideals m of R such that R/m ∼= K}(a1, . . . , ad) 7→ (x1 − a1, . . . , xd − ad).

Proof. We observe first that each ideal (x1 − a1, . . . , xd − ad) is a maximal ideal with residue fieldK, and that these ideals are distinct: if xi − ai, xi − a′i are in the same ideal for ai 6= a′i, then theunit ai − a′i is in the ideal, so it is not proper. To see that the map is surjective, let π : R� K bea surjective map. We have π(xi) ∈ K so (x1 − π(x1), . . . , xd − π(xd)) ⊆ m. The quotient by thisideal is already K, so (x1 − π(x1), . . . , xd − π(xd)) = m.

Theorem 2.6. Let K be a field, and R = K[x1, . . . , xd]/I be a finitely generated K-algebra. Thereare bijections

ZK(I) ↔ {maximal ideals m of R s.t. R/m ∼= K} ↔ HomK−alg(R,K)

(a1, . . . , ad) ∈ Z(I) ⊆ Kd 7→ (x1 − a1, . . . , xd − ad)m 7→ R� R/m ∼= K

Proof. In light of the previous proposition, we need to show that a ∈ ZK(I) ⇔ ma descends toa maximal ideal of R. This boils down to showing that a ∈ ZK(I) if and only if I ⊆ ma. Thisfollows from the fact that a ∈ ZK(I) ⇔ I ⊆ ker(evaluation at a) = ma. The second map is clearlybijective.

2.2. A QUICK WORD ABOUT TRANSCENDENCE BASES 21

2.2 A quick word about transcendence bases

We will need the notion of transcendence basis for a field extension shortly, and also later when wereturn to dimension theory.

Definition 2.7. Let K ⊆ L be an extension of fields. A transcendence basis for L over K is amaximal algebraically independent subset of L.

Observe that if {xλ}λ∈Λ is a transcendence basis for L over K, then we have a factorization

K ⊆ K({xλ}λ∈Λ) ⊆ L

where the first inclusion is purely transcendental , or isomorphic to a field of rational functions, andthe second inclusion is algebraic; if the latter were not algebraic, there would be an element of L tran-scendental overK({xλ}λ∈Λ), and we could use that element to get a larger algebraically independentsubset, contradicting the definition of transcendence basis. Conversely, if K ⊆ K({xλ}λ∈Λ) ⊆ Lwith the first inclusion purely transcendental and the second algebraic, {xλ}λ∈Λ is a transcendencebasis.

Here is the fact we will use momentarily.

Lemma 2.8. If A = {aλ}λ∈Λ is a field generating set for L over K, then A contains a transcendencebasis for L over K.

Consequently, any field extension admits a transcendence basis, and a finitely generated fieldextension admits a finite transcendence basis.

Proof. Observe first that if we have a nested union of algebraically independent sets, the union isalgebraically independent; if there were a nontrivial relation on some elements in the union, therewould be a nontrivial relation on finitely many, and so a relation in one of the members in the chain.By Zorn’s Lemma, we can then pick a maximal algebraically independent (over K) subset A′ of A.Any element of ArA′ must be algebraic over K(A′), and so L = K(A′)(ArA′) is algebraic overK(A′). By the observation above, A′ is a transcendence basis.

Here is a fact we will use later, whose proof we omit.

Theorem 2.9. Let K ⊆ L be an extension of fields. If X and Y are two transcendence bases forL over K, then |X| = |Y |.

This justifies the following definition.

Definition 2.10. The transcendence degree of a field extension L over K is the common size ofany transcendence basis for the extension.

Observe that the transcendence degree is a lower bound for the size of any field generating set.

29 Agosto

2.3 Maximal ideals of finitely generated algebras

Lemma 2.11 (Zariski’s Lemma). Let K ⊆ L be fields. If L is a finitely generated K-algebra, thenL is a finite dimensional K-vector space.


Proof. Let L = K[h1, . . . , hd]. Since L = K(h1, . . . , hd), we can choose a transcendence basis forL/K from among the h’s, and after reordering, we may assume that h1, . . . , hc = x1, . . . , xc forma transcendence basis, and hc+1, . . . , hd are algebraic over K ′ = K(x1, . . . , xc). Then L is integraland algebra-finite over K ′, hence module-finite. Thus, if c = 0, we are done. Suppose that c 6= 0;we will obtain a contradiction to complete the proof.

We can apply the Artin-Tate Lemma to K ⊆ K ′ ⊆ L to see that K ′ is algebra-finite over K. Inparticular, there are fi, gi in the polynomial ring K[x1, . . . , xc] such that K ′ = K[f1g1 , . . . ,

fcgc

]. This

implies that any element of K ′ can be written as a fraction with denominator (g1 · · · gc)n for somen. But, the element 1

g1···gc+1 ∈ K′ cannot be written this way; if so, we would have

v

(g1 · · · gc)n=

1

g1 · · · gc + 1,

for some v with g1 · · · gc - v (since the polynomial ring is a UFD). But, the equation g1 · · · gcv+ v =(g1 · · · gc)n contradicts this.

Corollary 2.12. Let K be a field, and R be a finitely generated K-algebra. For any maximal idealm of R, R/m is a finite extension of K.

In particular, if K is algebraically closed, R/m ∼= K.

Along similar lines, we have the following:

Exercise 2.13. If R is a finitely generated Z-algebra, and K = R/m is a residue field of R, thenK is finite.

Corollary 2.14 (Maximal ideals of f.g. K = K-algebras). Let K be an algebraically closed field,and S = K[x1, . . . , xd] be a polynomial ring. There is a bijection

Kd ↔ {maximal ideals m of S}

given by (a1, . . . , ad) 7→ (x1 − a1, . . . , xd − ad). If R is a finitely generated K-algebra, we can writeR = S/I for a polynomial ring S, and there is an induced bijection

ZK(I) ⊆ Kd ↔ {maximal ideals m of R}.

Theorem 2.15 (Nullstellensatz). Let K be an algebraically closed field, and R = K[x1, . . . , xd] bea polynomial ring. If I ⊂ R is an ideal (proper ideal!), then ZK(I) 6= ∅.

Proof. Let I ⊆ R be a proper ideal. Then, there is some maximal ideal I ⊆ m, and Z(m) ⊆ Z(I).We can write m = (x1−a1, . . . , xd−ad), so Z(m) = {(a1, . . . , ad)}; in particular, it is nonempty.

To attack the main question, we will need an observation on inequations. Observe that, if f(x)is a polynomial, f(a) 6= 0 if and only if there is a solution y = b ∈ K to yf(a)−1 = 0. In particular,a system of polynomial equations and inequations

f1(x) = 0, . . . , fm(x) = 0, g1(x) 6= 0, . . . , gn(x) 6= 0

has a solution x = a if and only if the system

f1(x) = 0, . . . , fm(x) = 0, y1g1(x)− 1 = 0, . . . , yngn(x)− 1 = 0

has a solution (x, y) = (a, b). In fact, this is equivalent to a system in one extra variable:

f1(x) = 0, . . . , fm(x) = 0, yg1(x) · · · gn(x)− 1 = 0.

2.3. MAXIMAL IDEALS OF FINITELY GENERATED ALGEBRAS 23

Theorem 2.16 (Strong Nullstellensatz). Let K be an algebraically closed field, and R = K[x1, . . . , xd]be a polynomial ring. Let I ⊆ R be an ideal. The polynomial f vanishes on ZK(I) if and only iffn ∈ I for some n ∈ N.

Proof. If fn ∈ I, and a ∈ ZK(I), then f(a) ∈ K satisfies f(a)n = 0 ∈ K. Since K is a field,f(a) = 0. Thus, f ∈ ZK(I) as well.

Suppose that f(x) vanishes along ZK(I). By the discussion above, this implies that ZK(I +(yf − 1)) = ∅, in a polynomial ring in one more variable. By the Medium Nullstellensatz, we seethat 1 ∈ IR[y] + (yf − 1). Write I = (g1(x), . . . , gm(x)), and

1 = r0(x, y)(1− yf(x)) + r1(x, y)g1(x) + · · ·+ rm(x, y)gm(x).

We can map y to 1/f to get

1 = r1(x, 1/f)g1(x) + · · ·+ rm(x, 1/f)gm(x)

in the fraction field of R[y]. Since each ri is polynomial, there is a largest negative power of foccurring; say that fn serves as a common denominator. We can multiply by fn to obtain (on theLHS) fn as a polynomial combination of the g’s (on the RHS).

Definition 2.17. The radical of an ideal I is the ideal√I := {f ∈ R | ∃n : fn ∈ I}. An ideal is

a radical ideal if I =√I.

To see that√I is an ideal, note that if fm, gn ∈ I, then

(f + g)m+n−1 =m+n−1∑i=0

(m+ n− 1

i

)f igm+n−1−i

= fm(fn−1 +

(m+ n− 1

1

)fn−2g + · · ·+

(m+ n− 1

n− 1

)gn−1

)+ gn

((m+ n− 1

n

)fm−1 +

(m+ n− 1

n+ 1

)fm−2g + · · ·+ gm−1

)∈ I,

and (rf)m = rmfm ∈ I.

Exercise 2.18. If R is a ring and I an ideal, then R/I is reduced (has no nonzero nilpotents) ifand only if I is a radical ideal.

In this terminology, the Strong Nullstellensatz asserts that, if K = K, ZK(I) ⊆ ZK(f) if andonly if f ∈

√I.

Observe that ZK(√J) = ZK(J) whether or not K is algebraically closed: “⊆” is clear, and if

fn(a) = 0, then f(a) = 0, so a ∈ ZK(J) and f ∈√J implies f(a) = 0, and the equality of sets

follows.

Definition 2.19. If K is an algebraically closed field, and X ⊆ Kn is a subset of the form X =ZK(I) for some (radical) ideal I ⊆ K[x1, . . . , xn], then we call X an affine variety, or a subvarietyof Kn.

Corollary 2.20. Let K be an algebraically closed field, and R = K[x1, . . . , xd] a polynomial ring.There is an order-reversing bijection between the collection of subvarieties of Kd and the collection


of radical ideals of R:

{subvarieties of Kd} ↔ {radical ideals I ⊆ R}

XI7→ {f ∈ R | X ⊆ ZK(f)}

ZK(I)Z←[ I.

In particular, for two ideals I, J , ZK(I) = ZK(J) if and only if√I =√J .

Proof. We check that I and Z are inverse operations on the specified sets; namely that Z(I(X)) =X for any subvariety X, and I(Z(J)) = J for any radical ideal J .

The Strong Nullstellensatz says that I(Z(J)) =√J for any ideal J , hence I(Z(J)) = J for a

radical ideal J .

Given X we can write X = ZK(J) for some radical ideal J . Then Z(I(X)) = Z(I(Z(J))) =Z(J) = X.

Definition 2.21. Let K be an algebraically closed field, and X = ZK(I) ⊆ Kd be a subvariety ofKd. The coordinate ring of X is the ring K[X] := K[x1, . . . , xd]/I(X).

Since K[X] is obtained from the polynomial ring on the ambient Kd by quotienting out byexactly those polynomials that are zero on X, we interpret K[X] as the ring of polynomial functionson X. Note that every reduced finitely generated K-algebra is a coordinate ring of some zerosetX, and conversely.

We now want to discuss maps on the set of maximal ideals. We prepare with a lemma.

Lemma 2.22. Let R ⊆ S be an integral extension.

1. Every nonzero element of S has a nonzero S-multiple in R.

2. If R is a field and S is a domain, then S is a field.

Proof. 1. Let s ∈ S. Take an integral equation of dependence for s over R with lowest degree:

sn + r1sn−1 + · · ·+ rn = 0.

WLOG, rn 6= 0, otherwise we could take an integral equation of lower degree. Rewriting, wehave rn = s(−sn−1 − r1s

n−1 − · · · − rn−1) ∈ R, as required.

2. Given r ∈ R r 0, there is some s ∈ R such that k = rs ∈ K r 0. Then sk−1 is an inversefor r.

Proposition 2.23. Let K be a field, and ϕ : R → S be a map of finitely generated K-algebras.Then, for any maximal ideal n of S, m = ϕ−1(n) is a maximal ideal of R.

Proof. The map K ⊆ S/n is module-finite, hence the intermediate extension K ⊆ R/ϕ−1(n) ismodule-finite as well. Since R/ϕ−1(n) ⊆ S/n, it is a domain. By the previous lemma, R/ϕ−1(n) isa field.

3 Septiembre

2.4. THE PRIME SPECTRUM OF A RING 25

2.4 The prime spectrum of a ring

We have seen that the set of maximal ideals in a (reduced) finitely generated algebra over analgebraically closed field is bijective to a subset of Kn for some n.

We can carry more of the geometric information on this set of maximal ideals by giving it atopology. The Zariski topology on Kn is the topology whose closed sets are zero sets of polynomialequations, equivalently, ZK(I) for the ideals I ⊆ K[x1, . . . , xn]. In light of the Nullstellensatz, wecan generalize this construction to all rings.

The maximal spectrum of a ring R, denoted Max(R), is the set of maximal ideals of R endowedwith the topology with closed sets given by VMax(I) := {m ∈ Max(R) | m ⊇ I} as I varies. Bythe Nullstellensatz, for polynomial rings S over an algebraically closed field K, this space Max(S)has a natural homeomorphism to Kn with its Zariski topology, and for an ideal I and S as above,Max(S/I) has a natural homeomorphism to ZK(I) ⊆ Kn with the subspace topology coming fromthe Zariski topology. Moreover, this is functorial: for any map of finitely generated K-algebras,there is an induced map on maximal ideals.

This is not quite the right notion to deal with general rings, for at least two reasons. First,there are many many interesting rings with only one maximal ideal! Second, we would like to havea geometric space that is assigned functorially to a ring, meaning that ring homomorphisms inducecontinuous maps of spaces (in the other direction). For the inclusion A = K[x, y] = K[x−1, y] intoB = K(x)[y] = K(x− 1)[y], what maximal ideal in A would we assign to (y) ⊆ B? How could oneof (x, y) or (x− 1, y) have a better claim than the other?

Definition 2.24. Let R be a ring. The prime spectrum, or spectrum of R is the set of primeideals of R, denoted Spec(R). It is naturally a poset, partially ordered by inclusion. We also endowit with the topology with closed sets V (I) := {p ∈ Spec(R) | p ⊇ I} for ideals I ⊆ R (and ∅ = V (I)is closed).

We will justify that this forms a topology shortly.We will illustrate posets with Hasse diagrams: if an element is below something with a line

connecting them, the higher element is ≥ the lower one.

Example 2.25. The spectrum of Z is, as a poset:

(2) (3) (5) (7) (11) · · ·

(0)

The closed sets are of the form V ((n)), which are the whole space when n = 0, the empty set withn = 1, and any finite union of things in the top row.

Example 2.26. Here are a few elements in C[x, y]:

· · · (x, y) (x− 1, y) (x− 1, y − 1) (x− 7π, y) (x− i√

2, y − 1) · · ·

· · · (x) (x− 1) (x2 − y3) (x2 + y2 + 1) (y) · · ·

(0)


Note that Max(R) is a subspace of Spec(R) (that may be neither closed nor open).

Proposition 2.27. Let R be a ring, and Iλ, J be ideals (possibly improper).

1. If I ⊆ J , then V (J) ⊆ V (I).

2. V (I) ∪ V (J) = V (I ∩ J) = V (IJ).

3.⋂λ V (Iλ) = V (

∑λ Iλ).

4. Spec(R) has a basis given by open sets of the form D(f) := Spec(R)r V (f).

5. Spec(R) is quasicompact.

Proof. 1. Clear.

2. To see V (I) ∪ V (J) ⊆ V (I ∩ J), just observe that if p ⊇ I or p ⊇ J , then p ⊇ I ∩ J . SinceIJ ⊆ I ∩ J , we have V (I ∩ J) ⊆ V (IJ). To show V (IJ) ⊆ V (I) ∪ V (J), if p 6⊇ I, J , letf ∈ I r p, and g ∈ J r p. Then fg ∈ IJ r p since p is prime.

3. Clear.

4. We can write any open set as the complement of V ({fλ}) =⋂λ V (fλ), which is the union of

D(fλ).

5. Given a sequence of ideals Iλ,if∑

λ Iλ = R, then 1 is in the sum on the left, and thus 1 canbe realized in such a sum over finitely many indices:

∑λ Iλ = Iλ1 + · · · + Iλt . Thus, if we

have a family of closed sets with empty intersection,

∅ =⋂λ

V (Iλ) = V (∑λ

Iλ) = V (Iλ1 + · · ·+ Iλt) = V (Iλ1) ∩ · · · ∩ V (Iλt),

so some finite subintersection is empty.

Definition 2.28 (Induced map on Spec). Given a homomorphism of rings ϕ : R → S, we obtaina map on spectra ϕ∗ : Spec(S)→ Spec(R) given by ϕ∗(p) = ϕ−1(p).

The key point is that the preimage of a prime ideal is also prime. We will often write p∩R forϕ−1(p), even if the map is not necessarily an inclusion.

We observe that this is not only an order-preserving map, but also is continuous: if U ⊆ Spec(R)is open, we have U = Spec(R)r V (I) for some ideal I; then for a prime q of S,

q ∈ (ϕ∗)−1(U) ⇔ q ∩R 6⊇ I ⇔ q 6⊇ IS ⇔ q ∈ Spec(S)r V (IS).

Example 2.29. Let π : R → R/I be surjective. Then the map π∗ : Spec(R/I) → Spec(R)corresponds to the inclusion of V (I) into Spec(R), since primes of R/I correspond to primes of Rcontaining I.

We can use the spectrum of a ring to give an analogue of the strong Nullstellensatz that is validfor any ring. To prepare for this, we need a notion that we will use later.

Definition 2.30. A subset W ⊆ R of a ring R is multiplicatively closed if 1 ∈W and a, b ∈W ⇒ab ∈W .

2.4. THE PRIME SPECTRUM OF A RING 27

Lemma 2.31. Let R be a ring, I an ideal, and W a multiplicatively closed subset. If W ∩ I = ∅,then there is a prime ideal p with p ⊇ I and p ∩W = ∅.

Proof. Consider the family of ideals {J | J ⊇ I, J ∩W = ∅}. This is nonempty, since it containsI, and has some maximal element a by a basic application of Zorn’s Lemma. We claim a is prime.Suppose f, g /∈ a. By maximality, a + (f) and a + (g) both have nonempty intersection with W ,so there exist r1f + a1, r2g + a2 ∈ W , with a1, a2 ∈ a. If fg ∈ a, then (r1f + a1)(r2g + a2) =r1r2fg + r1fa2 + r2ga1 + a1a2 ∈W ∩ a, a contradiction.

5 Septiembre

Proposition 2.32 (Spectrum analogue of strong Nullstellensatz). Let R be a ring, and I be anideal. For f ∈ R,

V (I) ⊆ V (f)⇐⇒ f ∈√I.

Equivalently,⋂

p∈V (I) p =√I.

Proof. First to justify the equivalence of the two statements we observe:

V (I) ⊆ V (f)⇔ f ∈ p for all p ∈ V (I)⇔ f ∈⋂

p∈V (I)

p.

We prove the latter formulation.(⊇): It suffices to show that p ⊇ I implies that p ⊇

√I. But, fn ∈ p implies f ∈ p, so this is

clear.(⊆): If f /∈

√I, consider the multiplicatively closed set W = {1, f, f2, f3, . . . }. We have

W ∩ I = ∅ by hypothesis. By the previous lemma, there is a prime p in V (I) that does notintersect W , and hence does not contain f .

The following corollary follows in exactly the same way as the analogous statement for subva-rieties of Kn, Corollary 2.20.

Corollary 2.33. Let R a ring. There is an order-reversing bijection

{closed subsets of Spec(R)} ↔ {radical ideals I ⊆ R}.

In particular, for two ideals I, J , V (I) = V (J) if and only if√I =√J .

Chapter 3

Exact functors, localization, flatness

3.1 Exact, left-exact, right-exact sequences

Definition 3.1. A sequence of maps of R-modules

· · · →M1α1−→M2

α2−→ · · · αt−1−−−→Mt → · · ·

is said to be exact if for all 1 < i < t, one has ker(αi) = im(αi−1) as submodules of Mi. We allowsequences that may or may not continue indefinitely in either direction.

For a while, we will be happy to focus on a few special cases.

Definition 3.2. A short exact sequence of R-modules is an exact sequence of the form

0→ Lα−→M

β−→ N → 0;

that is:

• ker(α) = 0; i.e., α is injective,

• coker(β) = 0; i.e., β is surjective, and

• ker(β) = im(α).

Since α is injective, L ∼= α(L). If we identify L with α(L), then the data of the short exactsequence above translates to N ∼= M/L, with α : L → M the inclusion map, and β : M → N toquotient map.

Definition 3.3. A right exact sequence of R-modules is an exact sequence of the form

Lα−→M

β−→ N → 0;

that is:

• coker(β) = 0; i.e., β is surjective, and


The data of a right exact sequence translates to N ∼= coker(α) = M/ im(α), with β : M → Nthe quotient map from the target onto the cokernel.

A special case of this is when M and N are free.

29

30 CHAPTER 3. EXACT FUNCTORS, LOCALIZATION, FLATNESS

Definition 3.4. A presentation of a module N is a right-exact sequence of the form

F1α−→ F0 → N → 0

with F1 and F0 free.

The image of the free basis of F0 in N is a generating set for N ; this is equivalent to surjectivity.The image of F1 in F0 gives the relations on those generators.

Remark 3.5. We say that a module is finitely presented or finitely presentable if it admits apresentation by finitely generated free modules. If R is Noetherian, then finitely generated andfinitely presented are equivalent.

Definition 3.6. A left exact sequence of R-modules is an exact sequence of the form

0→ Lα−→M

β−→ N ;

that is:

• ker(α) = 0; i.e., α is injective, and


The data of a left exact sequence translates to L ∼= ker(β), and α : L → M is the inclusion ofthe kernel into the source.

Observe that a sequence of maps 0 → L → M → N → 0 is short exact if and only if it is leftand right exact.

3.2 Hom and projectives

Definition 3.7. Let L,M,N be R-modules.

• The module of homomorphisms from M to N is

HomR(M,N) := {φ : M → N | φ is R-linear}.

The R-module structure is given by the rule r ·φ is the homomorphism m 7→ rφ(m) = φ(rm).

• If α : M → N is a module homomorphism, we define a map HomR(L,α) or α∗ fromHomR(L,M)→ HomR(L,N) by the rule

α∗(φ) = α ◦ φ;

i.e.,

α∗ : ( Lφ−→M ) 7→ ( L

α−→Mφ−→ N ).

• If α : M → N is a module homomorphism, we define a map HomR(α,L) or α∗ fromHomR(N,L)→ HomR(M,L) by the rule

α∗(φ) = φ ◦ α;

i.e.,

α∗ : ( Nφ−→ L ) 7→ ( M

α−→ Nφ−→ L ).

3.2. HOM AND PROJECTIVES 31

Thus, given a fixed R-module L, F (−) := HomR(L,−) is a rule that assigns to any R-module

M another R-module F (M), and to any homomorphism Mφ−→ N a homomorphism F (M)

F (φ)−−−→F (N). This (plus the fact that F takes the identity map to the identity map and compositions tocompositions) makes F a covariant functor from R-modules to R-modules.

Similarly, given a fixed R-module L, G(−) := HomR(−, L) is rule that assigns to any R-

module M another R-module G(M), and to any homomorphism G(M)φ−→ G(N) a homomorphism

G(N)G(φ)−−−→ G(M). This (with the same caveats as above) makes G a contravariant functor from

R-modules to R-modules. The covariant vs. contravariant bit refers to whether the directions ofmaps have changed.

Given maps Lα−→ L′ andM

β−→M ′, we likewise get a map HomR(L′,M)HomR(α,β)−−−−−−−→ HomR(L,M ′),

by combining the constructions above.

Example 3.8. HomR(R,M) ∼= M by φ 7→ φ(1), and under this isomorphism, Mα−→ N corresponds

to 1 7→ m 1 7→ α(m) under this isomorphism.

If I is an ideal, HomR(R/I,M) ∼= annM (I) by the same map: the image of 1 in R/I must mapto something killed by I, and there is a unique R-linear map that does this. The same recipe formaps as above holds. Thus, we can identify HomR(R/I,−) with the functor that sends modulesM to annM (I), and sends maps to their restrictions to these submodules.

Theorem 3.9. 1. A sequence of maps

0→ Lα−→M

β−→ N

is left-exact if and only if, for all R-modules X, the sequence

0→ HomR(X,L)α∗−→ HomR(X,M)

β∗−→ HomR(X,N)

is left-exact.

2. A sequence of maps

Lα−→M

β−→ N → 0

is right-exact if and only if, for all R-modules X, the sequence

0→ HomR(N,X)β∗−→ HomR(M,X)

α∗−→ HomR(L,X)

is left-exact.

Proof. Let 0→ Lα−→M

β−→ N be left-exact, and X be an R-module.

• α∗ is injective: if Xφ−→ L is nonzero, X

φ−→ Lα−→M is as well, since a nonzero element in the

image of φ goes to something nonzero in the composition.

• ker(β∗) = im(α∗): Xφ−→ M

β−→ N is zero if and only if im(φ) ⊆ ker(β) = im(α), whichhappens if and only if φ factors through L; i.e., φ ∈ im(α∗).

The other direction of the first part follows from the example above; we can use X = R.

Let Lα−→M

β−→ N → 0 be a right-exact sequence, and X be an R-module.


• β∗ is injective: if Nφ−→ X is nonzero, pick n ∈ N not in the kernel, and m ∈M that maps to

n. Then, the image of m under Mβ−→ N

φ−→ X is nonzero.

• ker(α∗) = im(β∗): Lα−→ M

φ−→ X is zero if and only if im(α) ⊆ ker(φ), which happens if andonly if φ descends to a map of the form N ∼= M/ im(α)→ X; i.e., φ ∈ im(α∗).

Let Lα−→ M

β−→ N → 0 be a sequence of maps, and suppose that it is exact after applyingHomR(−, X) for all X.

• β is surjective: if not, let X = N/ im(β). There is a nonzero projection map Nφ−→ X, but

Mβ−→ N

φ−→ X is zero, contradicting injectivity of β∗.

• ker(β) ⊇ im(α): Take X = N , and Nid−→ X. Since ker(α∗) ⊇ im(β∗), L

α−→Mβ−→ N

id−→ X =

Lα−→M

β−→ N is zero.

• ker(β) ⊆ im(α): Take X = M/ im(α), and Mφ−→ X the projection map. Since L

α−→Mφ−→ X

is zero, φ is in the image of β∗, so it factors through β. This is equivalent to the statedcontainment.

In short, HomR(X,−) is kernel-preserving, and HomR(−, X) turns cokernels into kernels.

10 Septiembre

Definition 3.10. An R-module is projective P if the functor HomR(P,−) sends short exact se-quences to short exact sequences. Equivalently, P is projective if M

α−→ N surjective implies

HomR(P,M)Hom(P,α)−−−−−−→ HomR(P,N) is surjective.

Free modules are always projective: HomR(R⊕Γ,M) ∼= M⊕Γ, and a map α gets sent to Γ copiesof itself. Every projective module is a direct summand of a free module: we can map a free moduleF onto P , and since HomR(P, F ) � HomR(P, P ), the identity map of P is in the image: theidentity factors through F . Every direct summand of a free module is projective as well (exercise!).

Example 3.11. Let R = S×T be a direct product of rings. Then S(= S×0) is a direct summandof R as an R-module: writing π for the projection, we have π((s, t)(s′, 0)) = π((ss′, 0)) = (ss′, 0) =(s, t)π((s′, 0)); but is not free, since a free generator would have annihilator zero, but every elementof S is annihilated by T . Thus not every projective module is free.

3.3 Tensor products and flat modules

Definition 3.12. Let M and N be R-modules. The tensor product of M and N is the moduleM ⊗R N generated by the set {m⊗ n | m ∈M,n ∈ N} with relations

(rm+m′)⊗ n− r(m⊗ n)−m′ ⊗ n m,m′ ∈M,n ∈ N, r ∈ Rm⊗ (rn+ n′)− r(m⊗ n)−m⊗ n′ m ∈M,n, n′ ∈ N, r ∈ R.

Observe that if M =∑

αRmα and N =∑

β Rnβ, then M ⊗RN =∑

α,βmα⊗nβ: any elementis a sum of elements of the form m⊗ n, and using the relations above, we can write any m⊗ n asa linear combination of the specified generators. In particular, the tensor product of two finitelygenerated modules is finitely generated.

Tensor products are characterized by a universal property.

3.3. TENSOR PRODUCTS AND FLAT MODULES 33

Definition 3.13. Let L, M , N be R-modules. A map ϕ : M ×N → L is bilinear over R if it islinear in both the M and N arguments: ϕ(rm+m′, n) = rϕ(m,n) +ϕ(m′, n) and ϕ(m, rn+ n′) =rϕ(m,n) + ϕ(m,n′). We write BilR(M,N ;L) for the set of bilinear maps M ×N → L.

Like Hom, Bil is an R-module, defined by the action of R on any of the inputs.

The following follows from the definition.

Proposition 3.14. Let L, M , N be R-modules. There is a bilinear map of R-modules ρ : M×N →M⊗RN such that for any bilinear map ϕ : M×N → L, there is a unique R-module homomorphismφ : M ⊗R N → L with ϕ = φ ◦ ρ. This map ρ is given by ρ(m,n) = m⊗ n.

Consequently, there is an R-module isomorphism HomR(M ⊗R N,L) ∼= BilR(M,N ;L).

Proof. With ϕ fixed, we observe that if an R-linear map φ with ϕ = φ ◦ ρ exists, then φ(m⊗ n) =ϕ(m,n) for all simple tensors m⊗n; since these generate, this determines the map completely, if itexists. To see the existence, we consider the map on the free module F generated by the symbolsm ⊗ n (with no relations) given by m ⊗ n 7→ ϕ(m,n). Using the biliearity relations on ϕ, we seethat all elements of the form (rm+m′)⊗n−r(m⊗n)−m′⊗n and m⊗(rn+n′)−r(m⊗n)−m⊗n′must map to zero, so the map factors through the tensor product.

Here are some consequences of this universal property.

Proposition 3.15. Let L,M,Mλ, N be R-modules.

1. R⊗RM ∼= M .

2. M ⊗R N ∼= N ⊗RM .

3. (L⊗RM)⊗R N ∼= L⊗R (M ⊗R N).

4. (⊕

λ∈ΛMλ)⊗R N ∼=⊕

λ∈Λ(Mλ ⊗R N).

5. If S is an R-algebra, and N is an S-module, M ⊗R (S ⊗S N) ∼= M ⊗R N .

Proof. We prove a few, and leave the rest for you to verify or check in a difference source.

For #1, there is a bilinear map R ×M → M given by (r,m) 7→ rm, which induces a mapR ⊗R M → M sending r ⊗m 7→ m. The map φ : M → R ⊗M given by r 7→ 1 ⊗m is R-linear:φ(rm + m′) = 1 ⊗ (rm + m′) = r(1 ⊗m) + (1 ⊗m′) = rφ(m) + φ(m′), so the maps are mutuallyinverse R-module homomorphisms.

For #2, there is a bilinear map M ×N → N ⊗RM given by (m,n) 7→ n⊗m, which induces amap M ⊗RN → N ⊗RM sending m⊗ n 7→ n⊗m. An inverse is constructed in a similar way.

Definition 3.16. If Lα−→ L′ is a map of R-modules, and M is another R-module, there is an R-

module homomorphism L⊗RMα⊗M−−−→ L′⊗RM given by (α⊗M)(l⊗m) = α(l)⊗m on simple tensors.

This is the map coming from the universal property applied to L×M α×idM−−−−→ L′ ×M → L′ ⊗RM .

Likewise, given two maps Lα−→ L′, M

β−→M ′, there is a map L⊗RMα⊗β−−−→ L′ ⊗RM ′.

Thus, if N is an R-module, −⊗R N is a covariant functor.

Theorem 3.17 (Hom-tensor adjointness). Let L,M,N be R-modules. There is an isomorphismHomR(L ⊗R M,N) ∼= HomR(L,HomR(M,N)). These isomorphisms are functorial in each argu-ment (i.e., are natural transformations of functors of each argument L,M,N).


Proof. (Sketch) We have that HomR(L ⊗RM,N) ∼= Bil(L,M ;N) via the rule (L ⊗RM → N) 7→(L ×M ρ−→ L ⊗R M → N). We claim that BilR(L,M ;N) ∼= HomR(L,HomR(M,N)) by the mapsending φ 7→ (x 7→ φ(x,−)); the inverse is ψ 7→ ((x, y) 7→ ψ(x)(y)).

Theorem 3.18. If a sequence of maps

Lα−→M

β−→ N → 0

is right-exact, then

L⊗R Xα⊗X−−−→M ⊗R X

β⊗X−−−→ N ⊗R X(→ 0)

is right-exact.

Proof. To see that the latter sequence is right-exact, we show that if we apply HomR(−, Y ) to it,it yields a left-exact sequence, for all R-modules Y . The sequence becomes

HomR(N ⊗X,Y )Hom(β⊗X,Y ) //

∼=��

HomR(M ⊗X,Y )Hom(α⊗X,Y ) //

∼=��

HomR(L⊗X,Y )

∼=��

HomR(N,HomR(X,Y ))Hom(β,Hom(X,Y ))// HomR(M,HomR(X,Y ))

Hom(α,Hom(X,Y ))// HomR(L,HomR(X,Y )).

It follows from left-exactness of Hom that the bottom row is left-exact, so the top row must be aswell.

Remark 3.19. Let S be an R-algebra. If M is an R-module, then S⊗RM is an S-module, where

S acts on the S-factor. If Mα−→ N is a map of R-modules, then S ⊗RM

S⊗α−−−→ S ⊗R N is a map ofS-modules. The functor S ⊗R − is called extension of scalars from R to S.

Observe that S ⊗ R⊕Λ ∼= S⊕Λ as S-modules. By right-exactness, it follows that extension ofscalars preserves presentations. That is, if

RmA−→ Rn →M → 0

is a presentation, then

SmA−→ Sn → S ⊗RM → 0

is a presentation.

Definition 3.20 (Flat module). An R-module N is flat if, for any inclusion of modules M ′ ↪→M ,the map M ′ ⊗R N → M ⊗R N is injective. Equivalently, N is flat if − ⊗R N preserves exactsequences.

If φ : R→ S is a map of rings, we say that S is a flat R-algebra or φ is a flat homomorphismif S is a flat R-module.

Lemma 3.21. Projective modules, and in particular, free modules, are flat.

Proof. First, if F is free, then α ⊗R F is a direct sum of copies of the map α, so is α is injective,then so is α⊗R F .

3.3. TENSOR PRODUCTS AND FLAT MODULES 35

Now, if P is projective, there are maps Pi−→ F

p−→ P that compose to the identity. Thus

P ⊗RNi⊗RN−−−−→ F ⊗RN

p⊗RN−−−−→ P ⊗RN is the identity, so P ⊗RNi⊗RN−−−−→ F ⊗RN is always injective.

For any injection M ′ →M , we have a commutative diagram

M ′ ⊗R F //M ⊗R F

M ′ ⊗R P //

OO

M ⊗R P

OO

and since the diagonal is injective, M ′ ⊗R P →M ⊗R P must be injective.

12 Septiembre

Lemma 3.22. If φ : R→ S is a ring homomorphism, and M is a flat R-module, then S ⊗RM isa flat S-module.

Proof. (S⊗RM)⊗SN ∼= M ⊗R (S⊗SN) ∼= M ⊗RN , and under this identification, (S⊗RM)⊗S αagrees with M ⊗R α.

Proposition 3.23 (Equational criterion). If M is flat, the following condition holds:For any r = (r1, . . . , rt) ∈ Rt and m = (m1, . . . ,mt) ∈ M t with r · m = 0, there ex-

ist sj = (s1j , . . . , stj) ∈ Rt for j = 1, . . . , a such that r · sj = 0 and n1, . . . , na ∈ M such thatm =

∑aj=1 sjnj.

Proof. Let M be flat. Given a t-tuple of elements r ∈ Rt, consider the exact sequence 0 → K →Rt

[r]−→ R, where K is the kernel. By assumption, 0→ K ⊗RM →M t [r]−→M is exact. That is,

r ·m = 0⇒ m ∈ ker([r])⇒ m ∈ K ⊗RM.

Thus, such an m can be written as∑

j kj ⊗ nj for some kj ∈ K and nj ∈M . Unpackaging thedefinitions gives the elements we want.

Example 3.24. Suppose that r is a nonzerodivisor on R: ry = 0 ⇒ y = 0 for y ∈ R. If M is aflat module, x is a nonzerodivisor on M . Indeed, rm = 0 implies that there are s1, . . . , sj ∈ R suchthat rsj = 0 for each j and m ∈ (s1, . . . , sj)M , but all sj = 0 by hypothesis, so m = 0.

Another useful fact:

Proposition 3.25 (Hom and flat base change). Let S be a flat R algebra, and M,N be twoR-modules. Suppose that M is finitely presented. Then

S ⊗R HomR(M,N) ∼= HomS(S ⊗RM,S ⊗R N)

s⊗ ϕ 7→ s(S ⊗ ϕ)

is an isomorphism.

Proof. When M is R or a finitely generated free module R⊕a, this is clear: both sides are isomorphicto (S ⊗R N)⊕a, and it is easy to see that the map above realizes this.

Now, take a presentationR⊕b → R⊕a →M → 0.


If we apply S ⊗R −, we obtain another right exact sequence; if we then apply HomS(−, S ⊗R N)to this presentation, we obtain a left-exact sequence

0→ HomS(S ⊗RM,S ⊗R N)→ HomS(S ⊗R R⊕a, S ⊗R N)→ HomS(S ⊗R R⊕b, S ⊗R N).

Likewise, if we apply HomR(−, N), we obtain a left exact sequence; if we then apply S ⊗R −,we obtain by flatness another left exact sequence

0→ S ⊗R HomR(M,N)→ S ⊗R HomR(R⊕a, N)→ HomR(R⊕b, N).

We then have a diagram

0 // HomS(S ⊗RM,S ⊗R N) // HomS(S ⊗R R⊕a, S ⊗R N) // HomS(S ⊗R R⊕b, S ⊗R N)

0 // S ⊗R HomR(M,N) //

OO

S ⊗R HomR(R⊕a, N) //

∼=

OO

S ⊗R HomR(R⊕b, N),

∼=

OO

where the vertical maps are given by the formula of the statement. We claim that the squarescommute. Indeed, given X

α−→ Y ,

HomS(S ⊗R Y, S ⊗R N)Hom(S⊗α,S⊗N) // HomS(S ⊗R X,S ⊗R N)

S ⊗R HomR(Y,N)S⊗Hom(α,N) //

OO

S ⊗R HomR(X,N)

OO,

an element s⊗ϕ in the bottom left goes ↑ to s ·(S⊗ϕ) and then→ to s ·(S⊗(ϕ◦α)), whereas s⊗ϕgoes → to s⊗ (ϕ ◦ α) and then ↑ to s · (S ⊗ (ϕ ◦ α)). It then follows that there is an isomorphismin the first vertical map in the previous diagram.

We want to briefly note a definition related to flatness.

Definition 3.26 (Faithfully flat module). A module N over a ring R is faithfully flat if Lα−→ M

is injective if and only if L⊗R Nα⊗N−−−→M ⊗R N is injective.

We can talk of faithfully flat algebras / morphisms as with flat algebras / morphisms.

Proposition 3.27. Let M be a faithfully flat R-module, and S an R-algebra. Then M ⊗R S isfaithfully flat over S. In particular, if R is a K-algebra, and L is an extension field of K, thenR⊗K L is faithfully flat over R.

3.4 Localization

Recall the notion of multiplicative set.Our three most important classes of examples are the first three below.

Example 3.28. Let R be a ring.

1. For any f ∈ R, the set W = {1, f, f2, f3, . . . } is a multiplicative set.

2. If p ⊆ R is a prime ideal, the set W = Rrp is multiplicative: this is an immediate translationof the definition.

3.4. LOCALIZATION 37

3. The set of nonzerodivisors in R—elements that are not zerodivisors—forms a multiplicativelyclosed subset.

4. An arbitrary intersection of multiplicatively closed subsets is multiplicatively closed. In par-ticular, for any family of primes {pλ}, the set Rr

⋃λ pλ is multiplicatively closed.

Definition 3.29 (Localization of a ring). Let R be a ring, and W be a multiplicative set with0 /∈W . The localization of R at W is the ring

W−1R :={ rw

∣∣∣ r ∈ R,w ∈W} / ∼where ∼ is the equivalence relation

r

w∼ r′

w′if ∃u ∈W : u(rw′−r′w) = 0. The operations are given

byr

v+s

w=rw + sv

vwand

r

v

s

w=

rs

vw.

There is a canonical ring homomorphism R→W−1R that sends r 7→ r1 .

Note that we write elements in W−1R in the form r/w even though they are equivalence classesof such expressions.

Observe that if R is a domain, the equivalence relation simplifies to rw′ = r′w, so R ⊆W−1R ⊆Frac(R), and in particular W−1R is a domain too.

In the localization of R at W , every element of W becomes a unit. The following universalproperty says roughly that W−1R is the smallest R-algebra in which every element of R is a unit.

Proposition 3.30. Let R be a ring, and W a multiplicative set with 0 /∈W . Let S be an R-algebrain which every element of W is a unit. Then there is a unique homomorphism α such that thefollowing diagram commutes:

R //

��

W−1R

α{{

S

where the vertical map is the structure homomorphism and the horizontal map is the canonicalhomomorphism.

Example 3.31 (Most important localizations). Let R be a ring.

1. For f ∈ R and W = {1, f, f2, f3, . . . }, we usually write Rf for W−1R.

2. For p ⊂ R prime, we generally write Rp for (Rr p)−1R.

3. When W is the set of nonzerodivisors on R, we call W−1R the total ring of fractions of R.When R is a domain, this is just the fraction field of R.

We state an analogous definition for modules, and for module homomorphisms.

Definition 3.32. Let R be a ring, W be a multiplicative set, and M an R-module. The localizationof M at W is the W−1R-module

W−1M :={mw

∣∣∣ m ∈M,w ∈W}/ ∼


where ∼ is the equivalence relationm

w∼ m′

w′if ∃u ∈ W : u(mw′ −m′w) = 0. The operations are

given bym

v+n

w=mw + nv

vwand

r

v

m

w=rm

vw.

If Mα−→ N is an R-module homomorphism, then there is a W−1R-module homomorphism

W−1MW−1α−−−−→W−1N given by the rule W−1α(m/w) = α(m)/w.

We will use the notations Mf and Mp analogously to Rf and Rp.

To understand localizations of rings and modules, we will want to understand better how theyare built from R.

Lemma 3.33. Let M be an R-module, and W a multiplicative set. The class

m

w∈W−1M is zero ⇐⇒ ∃v ∈W : vm = 0 ⇐⇒ annR(m) ∩W 6= ∅.

Note in particular this holds for w = 1.

Proof. For the first equivalence, we compute: mw = 0

1 in W−1M if and only if ∃v ∈ W such that0 = v(1m−0w) = vm. The second equivalence just comes from the definition of the annihilator.

Remark 3.34. It follows from this lemma that if α : N →M is injective, then W−1α : W−1N →W−1M is as well. Indeed, if α is injective, then

0 = W−1α(n/w) = α(n)/w ⇒ ∃u ∈W : 0 = uα(n) = α(un)⇒ un = 0⇒ n/w = 0.

We want to collect one more lemma for later.

Lemma 3.35. Let M be a module, and N1, . . . , Nt be a finite collection of submodules. Let W bea multiplicative set. Then,

W−1(N1 ∩ · · · ∩Nt) = W−1N1 ∩ · · · ∩W−1Nt ⊆W−1M.

Proof. The containment “⊆” is clear. An element of the RHS is of the form n1w1

= · · · = ntwt

; we canfind a common denominator to realize this in the LHS.

17 Septiembre

Theorem 3.36 (Flatness of localization). Let R be a ring, and W a multiplicative system. Then

1. W−1R⊗RM ∼= W−1M as W−1R-modules, and W−1R⊗α corresponds to W−1α under theseisomorphisms.

2. W−1R is flat over R.

3. W−1(−) is an exact functor; i.e., it sends exact sequences to exact sequences.

3.4. LOCALIZATION 39

Proof. 1. The bilinear map W−1R×M →W−1M given by (r/w,m) 7→ rm/w induces a map ψfrom the tensor product that is clearly surjective. For an inverse map, set φ(m/w) = 1/w⊗m.To see this is well-defined, suppose m/w = m′/w′, so ∃v ∈ W such that v(mw′ −m′w) = 0.Then,

φ(m/w)− φ(m′/w′) = 1/w ⊗m− 1/w′ ⊗m′.

We can multiply through by vww′/vww′ to get

vw′

vww′⊗m− vw

vww′⊗m′ = 1

vww′⊗ v(mw′ −m′w) = 0.

To see this is a homomorphism, we note that

φ(m

w+m′

w′) = φ(

mw′ +m′w

ww′) =

1

ww′⊗ (mw′ +m′w) =

1

ww′⊗mw′ + 1

ww′⊗m′w

=w′

ww′⊗m+

w

ww′⊗m′ = 1

w⊗m+

1

w′⊗m′ = φ(

m

w) + φ(

m′

w′),

and

φ(rm

w) =

1

w⊗ rm = r(

1

w⊗m) = rφ(

m

w).

The composition φ ◦ ψ sends r/w ⊗ m 7→ rm/w 7→ 1/w ⊗ rm = r/w ⊗ m; since it is theidentity on simple tensors and additive, it is the identity.

For the claim about maps, we need to see that, forMα−→ N , we have to check ψN ◦ (W−1R⊗ α) =

W−1α ◦ ψM . Indeed,

(ψN ◦ (W−1R⊗ α))(r

w⊗m) = ψN (

r

w⊗ α(m)) =

rα(m)

w

=α(rm)

w= W−1α(

rm

w) = (W−1α ◦ ψM )(

r

w⊗m).

2. This follows from the earlier observation that W−1(−) preserves injective maps.

3. This is immediate from part (2).

Corollary 3.37 (Hom and localization). Let R be a Noetherian ring, W be a multiplicative set,M be a finitely generated R-module, and N an arbitrary R-module. Then,

HomW−1R(W−1M,W−1N) ∼= W−1 HomR(M,N).

In particular, if p is prime,

HomRp(Mp, Np) ∼= HomR(M,N)p.

Proposition 3.38. Let M be an R-module, and W a multiplicative set. Set MW = {m ∈M | ∃w ∈W : wm = 0} and M = M/MW . Then,

W−1M =⋃w∈W

1

wM,

where each 1wM is an isomorphic copy of M as an R-module.


Proof. First, MW is a submodule of M and consists of the elements that go to zero in the local-ization. We have W−1M ∼= W−1(M/MW ) ∼= W−1M

W−1MW

∼= W−1M , so we can replace M by the

W -torsionfree module M . It is clear that every element of W−1M has the form on the RHS.The map M → 1

wM sending m 7→ mw is clearly R-linear, and is injective by the W -torsion free

assumption.

Proposition 3.39. Let W be multiplicatively closed in R.

1. If I is an ideal, then W−1I ∩R = {r ∈ R | ∃w ∈W : wr ∈ I}.

2. If p is prime and W ∩ p = ∅, then W−1p = p(W−1R) is prime.

3. The map Spec(W−1R)→ Spec(R) is injective, with image {p ∈ Spec(R) | p ∩W = ∅}.

Proof. 1. Since W−1(R/I) ∼= W−1R/W−1I, we have ker(R → W−1(R/I)) = R ∩W−1I. Theequality is then clear.

2. First, since W ∩p = ∅, and p is prime, we know that no element of W kills 1 = 1+p in R/p, so1/1 is nonzero in W−1(R/p). Thus, W−1R/W−1p ∼= W−1(R/p) nonzero, and a localizationof a domain, hence is a domain. Thus, W−1p is prime.

3. First, by part (2), the map p 7→ W−1p, for S = {p ∈ Spec(R) | p ∩W = ∅} sends primes toprimes. We claim that

Spec(W−1R) ↔ S

q 7→ q ∩RW−1p = p(W−1R) ←[ p

is a pair of mutually inverse maps.

To see this, first note that if J is an ideal of W−1R, then J = (J ∩ R)W−1R. Indeed, ifJ = (a1/w1, . . . , at/wt), then J = (a1/1, . . . , at/1), since each generator was replaced by aunit multiple. Second, if W ∩ p = ∅, then using part (1) and the definition of prime, we havethat p = W−1p ∩R.

Corollary 3.40. Let R be a ring and p be a prime ideal. The map R → Rp induces a map onspectra corresponding to the inclusion

{q ∈ Spec(R) | q ⊆ p} ↪→ Spec(R).

Definition 3.41. Let φ : R→ S be a ring homomorphism, and p ∈ Spec(R). We call the ring

κφ(p) := (Rr p)−1(S/pS)

the fiber ring of φ over p.

The point of this definition is the following.

Lemma 3.42. Let φ : R → S be a ring homomorphism, and p ∈ Spec(R). Spec(κφ(p)) ∼=(φ∗)−1(p), the set of primes that contract to p.

Proof. Consider the maps S → S/pS → (R r p)−1(S/pS). For the first map, the map on spectracan be identified with the inclusion of V (pS) into Spec(S). For the second, it can be identifiedwith the inclusion of the set of primes that do not intersect Rr p, i.e., those whose contraction iscontained in p. Put together, this is the set of primes that contract to p.

Chapter 4

Decomposition of ideals

4.1 Minimal primes and support

We will consider a few ways of decomposing ideals into pieces, in three ways with increasing detail.The first is the most directly geometric: for any ideal I in a Noetherian ring, we aim to write V (I)as a finite union of V (pi) for prime ideals pi.

Definition 4.1. The primes that contain I and are minimal with the property of containing I arecalled the minimal primes of I. That is, the minimial primes of I are the minimal elements ofV (I). We write Min(I) for this set.

Lemma 4.2. Let R be a ring, and I an ideal. Every prime p that contains I contains a minimalprime of I. Consequently,

√I =

⋂p∈V (I) p =

⋂p∈Min(I) p.

Proof. This follows from Zorn’s lemma. We just need to check that the intersection of a descendingchain of prime ideals that contain I is prime and contains I. These are both trivial. The firstequality in the “consequently” we already know; the second is basic set theory.

Remark 4.3. If p is prime, then Min(p) = {p}. Also, we have Min(I) = Min(√I) (since V (I) =

V (√I)).

As a special case, the nilpotent elements of a ring R are exactly the elements in every minimalprime of R (equivalently, every minimal prime of the zero ideal).

Lemma 4.4. Given an expression I = p1 ∩ · · · ∩ pn with pi 6⊆ pj for each i, j, we have Min(I) ={p1, . . . , pn}.

Proof. This boils down to the claim that if q is prime, and q ⊇ (p1 ∩ · · · ∩ pn) then q containssome pi. But, if q 6⊇ pi for each i, there are elements fi ∈ pi r q, and the product f1 · · · fn ∈(p1 ∩ · · · ∩ pn)r q.

Theorem 4.5. Let R be a Noetherian ring. Then any ideal I has finitely many minimal primes,and thus, we can write √

I = p1 ∩ · · · ∩ pn.

Proof. Let S = {ideals I ⊆ R | Min(I) is infinite}, and suppose, to obtain a contradiction, thatS 6= ∅. By Noetherianity, S has a maximal element J . Clearly, J is not prime (for Min(J) wouldbe a singleton), but J is radical, since Min(J) = Min(

√J). Partitioning the minimal primes of

J into two nonempty sets we can write J = J1 ∩ J2, where J1 is the intersection of some of the

41

42 CHAPTER 4. DECOMPOSITION OF IDEALS

primes, and J2 the intersection of the others, and J 6= J1, J2, since V (J1), V (J2) $ V (J) and theideals are radical. By choice of J , we know that Min(J1) and Min(J2) are finite. We then haveJ1 = p1 ∩ · · · ∩ pa and J2 = pa+1 ∩ · · · ∩ pb, where the p′cs are minimal primes, so J = p1 ∩ · · · ∩ pb.By the previous lemma, we see that Min(J) is a subset of {p1, . . . , pb}, so is finite.

19 Septiembre

We now can describe the relationship between the poset structure of Spec(R) and the topology.

Proposition 4.6. Let R be a ring, and X = Spec(R).

1. The poset structure on X can be recovered from the topology by the rule: p ⊆ q (p ≤ q) ⇔q ∈ {p}.

2. If R is Noetherian, the topology on X can by recovered from the poset structure by the rule

Y ⊆ X is closed ⇔ ∃p1, . . . , pn ∈ X : Y = {q ∈ X | pi ⊆ q for some i}.

Proof. 1. We calculate the closure of a point. We have {p} =⋂

p∈V (I) V (I). Note that p ∈ V (I)

implies I ⊆ p, which implies V (p) ⊆ V (I). It follows that {p} = V (p), and thus the claim.

2. If Y is closed, we have Y = V (I) = V (√I) = V (p1 ∩ · · · pn) = V (p1) ∪ · · · ∪ V (pn). For the

converse, we can work backwards.

We now wish to understand modules in a similar way.

Definition 4.7. If M is an R-module, the support of M is Supp(M) := {p ∈ Spec(R) | Mp 6= 0}.

Proposition 4.8. If R is a ring, and M a finitely generated module, then Supp(M) = V (annR(M)).In particular, Supp(R/I) = V (I).

Proof. LetM =∑

iRmi. We have that annR(M) =⋂i annR(mi), so V (annR(M)) =

⋃i V (annR(mi));

we are using finiteness here. Also, observe that Supp(M) =⋃i Supp(Rmi): the containment ⊇ is

clear since (Rmi)p ⊆Mp, while ⊆ follows from the fact that the images of mi’s in Mp generate Mp

for each p. Thus, we reduce to the case of Rmi, and this follows from fact that mi/1 = 0 in Mp ifand only if (Rr p) ∩ annR(mi) 6= ∅, which happens if and only if annR(mi) 6⊆ p.

The finite generating hypothesis is necessary!

Example 4.9. Let K be a field, and R = K[x]. Take M = Rx/R =⊕

i>0Kx−i. With this

K-vector space structure, the action is given by multiplication in the obvious way, then killing anynonnegative degree terms.

On one hand, Supp(M) = {(x)}. Indeed, any element of M is killed by a large power of x, soW−1M = 0 if x ∈W , so Supp(M) ⊆ {(x)}.

On the other hand, the annihilator of the class of x−n is xn, so annR(M) ⊆⋂n∈N(xn) = 0.

Example 4.10. Let R = C[x], and M =⊕

n∈ZR/(x− n).On the one hand, Supp(M) = {(x− n) | n ∈ Z}. Indeed, Mp =

⊕n∈Z(R/(x− n))p, so

Supp(M) =⋃n∈Z

Supp(R/(x− n)) =⋃n∈Z

V ((x− n)) = {(x− n) | n ∈ Z}.

On the other hand, AnnR(M) =⋂n∈Z annR(R/(x− n)) =

⋂n∈Z(x− n) = 0.

Note that the support is not even closed.

4.2. ASSOCIATED PRIMES AND PRIME FILTRATIONS 43

Proposition 4.11. Let R be a ring, L,M,N be modules, and m ∈M .

1. m = 0 ⇔ m/1 = 0 in Mp for all p ∈ SpecR ⇔ m/1 = 0 in Mm = 0 for all m ∈ MaxR.

2. If 0→ L→M → N → 0 is exact, then Supp(L) ∪ Supp(N) = Supp(M).

3. M = 0 ⇔ Mp = 0 for all p ∈ SpecR ⇔ Mm = 0 for all m ∈ MaxR.

Proof. 1. The implications ⇒ are clear. To show the last implies the first, we show the comtra-positive. If m 6= 0, its annihilator is a proper ideal, which is contained in a maximal ideal, soV (annRm) = Supp(Rm) contains a maximal ideal.

2. For any p, we have 0 → Lp → Mp → Np → 0 is exact. If p ∈ Supp(L) ∪ Supp(N), thenLp or Np is nonzero, so Mp must be. On the other hand, if p /∈ Supp(L) ∪ Supp(N), thenLp = Np = 0, so Mp = 0.

3. The implications ⇒ are clear. To show the last implies the first, we show the comtrapositive.If m 6= 0, consider Rm ⊆ M . By part (1), there is a maximal ideal in Supp(Rm), and bypart (2), this maximal ideal is in Supp(M) as well.

4.2 Associated primes and prime filtrations

We now refine our decomposition of ideals/modules.

Definition 4.12. Let R be a ring, and M a module. We say that p ∈ Spec(R) is an associatedprime of M if p = annR(m) for some m ∈ M . Equivalently, p is associated to M if there is aninjective homomorphism R/p ↪→M . We write AssR(M) for the set of associated primes of M .

If I is an ideal, by the associated primes of I we (almost always) mean the associated primesof R/I; but we’ll try to write AssR(R/I).

Lemma 4.13. If p is prime, AssR(R/p) = {p}.

Proof. For any nonzero r ∈ R/p, we have annR(r) = {s ∈ R | rs ∈ p} = p by definition of primeideals.

Lemma 4.14. If R is Noetherian, and M an arbitrary R-module, then

1. Ass(M) = ∅⇔M = 0, and

2.⋃

p∈Ass(M)

p = {zerodivisors on M} := {r ∈ R | rm = 0 for some m ∈M r {0}}.

Additionally, if R and M are (Z-)graded and M 6= 0, M has an associated prime that is homoge-neous.

Proof. (⇐) is clear in part 1 (and certainly doesn’t require Noetherian).The interesting direction of 1 and part 2 will both follow from the fact that any ideal of the form

annR(m) is contained in an associated prime; such a prime will certainly exist, and if something isa zerodivisor, it must belong to an associated prime.

Any element in the (nonempty) set {annR(m) | m ∈M r0} is contained in a maximal element,by Noetherianity. Let I = ann(m) be such an element, and let rs ∈ I, s /∈ I. Clearly ann(sm) ⊇ann(m), and equality holds by maximality. Then, (rs)m = r(sm) = 0, so r ∈ ann(sm) = ann(m) =I. Thus, I is prime.


For the graded case, replace the set about with the annihilators of homogeneous elements. Suchannihilator is homogeneous, since if m is homogeneous, and fm = 0, writing f = fa1 + · · ·+ fab asa sum of homogeneous pieces, then 0 = fm = fa1m + · · ·+ fabm is a sum of elements of differentdegrees, so faim = 0 for each i.

The same argument above works if we take {annR(m) | m ∈ M r 0 homogeneous }, using thefollowing lemma.

Lemma 4.15. If R is Z-graded, an ideal with the property

∀r, s ∈ R homogeneous rs ∈ I ⇒ r ∈ I or s ∈ I

is prime.

Proof. We need to show that this property implies that a, b ∈ R, ab ∈ I implies a ∈ I or b ∈ I. Weinduce on the number of nonzero homogeneous components of a plus the number of nonzero homo-geneous components of b. The base case is when this is two, coming directly from the hypotheses.Otherwise, write a = a′ + am, b = b′ + bn, where am, bn are the largest homogeneous componentsof a, b respectively. We have ab = (a′b′ + amb

′ + bna′) + ambn, where ambn is either the largest

homogeneous component of ab or else it is zero. Either way, ambn ∈ I, so am ∈ I or bn ∈ I; WLOGam ∈ I. Then ab = a′b + amb, and ab, amb ∈ I, so a′b ∈ I, and the total number of homogeneouspieces is smaller, so by induction, either a′ ∈ I so that a ∈ I, or else b ∈ I.

Lemma 4.16. If 0→ L→M → N → 0 is exact, then Ass(L) ⊆ Ass(M) ⊆ Ass(L) ∪Ass(N).

Proof. If R/p ↪→ L, then composition with the inclusion L ↪→ M gives R/p ↪→ M . Let p ∈Ass(M) r Ass(L), and let p = ann(m). Now, every submodule of Rm consists of 0 and elementswith annihilator p, so Rm∩L = 0. Thus, Rm ⊆M bijects onto its image in N in the map M → N ,so R/p ↪→ N .

24 Septiembre

We will need a bit of notation for graded modules to help with the next statement.

Definition 4.17. Let R and M be T -graded, and t ∈ T . The shift of M by t is the graded R-module M(t) with graded pieces M(t)i := Mt+i. This is isomorphic to M as an R-module withoutthe grading.

Theorem 4.18. If R is a Noetherian ring, and M is a finitely generated module, then there existsa filtration of M

M = Mt %Mt−1 %Mt−2 % · · · %M1 %M0 = 0

such that Mi/Mi−1∼= R/pi for primes pi ∈ Spec(R). Such a filtration is called a prime filtration

of M .If additionally R and M are Z-graded, there is a filtration as above with Mi/Mi−1

∼= (R/pi)(ti)as graded modules for homogeneous primes pi ∈ Spec(R) and integers ti.

Proof. If M 6= 0, then M has an associated prime, so there is an injection M1∼= R/p1 ↪→ M . If

M/M1 6= 0, it has an associated prime, so there is an M2 ⊆M such that M2/M1∼= R/p2 ↪→ R/M1.

Continuing this process, we get a strictly ascending chain of submodules of M where the successivequotients are of the form R/pi. If we do not have Mt = M for some t, then we get an infinitestrictly ascending chain of submodules of M , which contradicts that M is a Noetherian module.

4.2. ASSOCIATED PRIMES AND PRIME FILTRATIONS 45

In the graded case, if pi is the annihilator of an element mi of degree ti, we have a degree-preserving map (R/pi)(ti) ∼= Rmi sending the class of 1 to mi. The rest of the proof is thesame.

Prime filtrations often allow us to reduce statements about finitely generated modules to state-ments about quotient domains of R: modules of the form R/p for primes p.

Corollary 4.19. If R is a Noetherian ring, and M is a finitely generated module, and

M = Mt %Mt−1 %Mt−2 % · · · %M1 %M0 = 0

is a prime filtration of M with Mi/Mi−1∼= R/pi then

AssR(M) ⊆ {p1, . . . , pt}.

Consequently,

• AssR(M) is finite.

• If M is graded, AssR(M) is a finite set of homogeneous primes.

Proof. For each i, we have Ass(Mi) ⊆ Ass(Mi−1) ∪ Ass(Mi/Mi−1) = Ass(Mi−1) ∪ {pi} so that,inductively, Ass(Mi) ⊆ {p1, . . . , pi}. The consequences follow from the previous theorem.

Example 4.20. Any subset X ⊆ Spec(R) (for any R) can be realized as Ass(M) for some M : takeM =

⊕p∈X R/p. Of course, if X is infinite, M is not finitely generated.

Example 4.21. If R is not Noetherian, then there may be modules (or ideals even) with noassociated primes. Let R =

⋃n∈NCJx1/nK be the ring of nonnegatively-valued Puiseux series. We

claim that R/(x) is a cyclic module with no associated primes (i.e., the ideal (x) has no associatedprimes). First, observe that any element of R can be written as a unit times xm/n for somem,n, so any associated prime must be an annihilator of xm/n + (x) for some m ≤ n. We haveann(xm/n + (x)) = (x1−m/n), which is not prime, since (x1/2−m/2n)2 ∈ (x1−m/n), but x1/2−m/2n.

We will need that associated primes localize.

Theorem 4.22 (Associated primes localize in Noetherian rings). Let R be a Noetherian ring, W amultiplicative set, and M a module. Then AssW−1R(W−1M) = {W−1p | p ∈ AssR(M), p∩W = ∅}.

Proof. (⊇): Given p ∈ AssR(M), p ∩ W = ∅, we have that W−1p is a prime (proper ideal) inW−1R. Then W−1R/W−1p ∼= W−1(R/p) ↪→W−1M by exactness, so it is associated.

(⊆): If W−1p is associated to W−1M , there is an embedding

W−1(R/p)(∼= W−1R/W−1p)i↪→W−1M.

By the Noetherian hypothesis, sinceR/p is finitely generated, Hom localizes: W−1 HomR(R/p,M) ∼=HomW−1R(W−1R/W−1p,W−1M), so there is some R/p

i′−→ M and w ∈ W such that i = w−1 ·W−1i′. Let K = ker(i′). Since W−1K = ker(W−1i) = 0 by exactness, every element of K is killedby something in W . But, K ⊆ R/p, so elements of W act as nonzerodivisors on K. Hence, K = 0.Thus, R/p injects into M , so p ∈ AssR(M).

Corollary 4.23. Let R be Noetherian, and M be an R-module.


1. SuppR(M) =⋃

p∈AssR(M) V (p).

2. If M is finitely generated, then Min(annR(M)) ⊆ AssR(I). In particular, Min(I) ⊆ AssR(R/I).

Proof. 1. (⊇): Let p ∈ AssR(M) and let p = annR(m) for m ∈ M . Let q ∈ V (p). We have0→ R/p

m−→M exact, and thus 0→ (R/p)q →Mq is exact, with (R/p)q nonzero, so Mq 6= 0.

(⊆): Suppose that q is not in the right-hand side in the equation above, so that q does notcontain any associated prime of M . Then there is no associated prime of M that does notintersect Rr q, so AssRq(Mq) = ∅, so Mq = 0.

2. We have that V (annR(M)) = SuppR(M) =⋃

p∈AssR(M) V (p), so the minimal elements ofboth sets agree. In particular, the right hand side has the minimal primes of annR(M) asminimal elements, and they must be associated primes of M , or else this would contradictminimality.

Definition 4.24. If I is an ideal, then an associated prime of I that is not a minimal prime of Iis called an embedded prime of I.

25 Septiembre

We take a quick detour to introduce an important lemma.

Lemma 4.25 (Prime avoidance). Let R be a ring, I1, . . . , In, J be ideals, and suppose that Ii isprime for i > 2 (at most two are not prime).

If J 6⊆ Ii for all i, then J 6⊆⋃i Ii; equivalently, if J ⊆

⋃i Ii, then J ⊆ Ii for some i.

Moreover, if R is N-graded, and all of the ideals are homogeneous, all Ii are prime, and J 6⊆ Iifor all i, then there is a homogeneous element in J r

⋃i Ii.

Proof. We proceed by induction on n. If n = 1, there is nothing to show.

By induction hypothesis, we can find elements ai ∈ J r⋃j 6=i Ij for each i. If some ai /∈ Ii,

we are done, so suppose that ai ∈ Ii for each i. Consider a = an + a1 · · · an−1. This belongs toJ . If a ∈ Ii for i < n, then, since a1 · · · an−1 = ai(a1 · · · ai · · · an−1) ∈ Ii, we also have an ∈ Ii, acontradiction. If a ∈ In, then, since an ∈ In, we also have a1 · · · an−1 ∈ In. If n = 2, this saysa1 ∈ I2, a contradiction. If n > 2, then In is prime, so one of a1, . . . , an−1 ∈ In, a contradiction.

If all Ii are homogeneous and prime, we proceed as above, replacing an and a1, . . . , an−1 withsuitable powers (e.g., |a1| + · · · + |an−1| and |an| each, respectively) so that an + a1 · · · an−1 ishomogeneous. The primeness assumption guarantees that noncontainments in ideals is preserved.

Corollary 4.26. Let I be an ideal and M a finitely generated module over a Noetherian ring R.If I consists of zerodivisors on M , then Im = 0 for some m ∈M .

Proof. We have that I ⊆⋃

p∈Ass(M)(p). By the assumptions, this is a finite set of primes. By primeavoidance, I ⊆ p for some p ∈ Ass(M). That is I ⊆ annR(m) for some m ∈M .

4.3. PRIMARY DECOMPOSITION 47

4.3 Primary decomposition

We refine our decomposition theory once again.

Definition 4.27. We say that an ideal is primary if xy ∈ I ⇒ x ∈ I or y ∈√I. We say that an

ideal is p-primary if I is primary and√I = p.

We observe that a primary ideal has a prime radical: if a is primary, and xy ∈√a, then

xnyn ∈ a for some n. If y /∈√a, then we must have xn ∈ a, so x ∈

√a. Thus, every primary ideal

a is√a-primary.

Example 4.28. 1. If R is a UFD, a principal ideal is primary if and only if it is generated by apower of a prime element. Indeed, if a = fn, with f irreducible, then xy ∈ (fn)⇔ fn|xy ⇔fn|x or f |y ⇔ x ∈ (fn) or y ∈

√(fn) = (f). Conversely, if a is not a prime power, then

a = gh, for some g, h nonunits with no common factor, then take gh ∈ (a) but g /∈ a andh /∈

√(a).

2. In R = C[x, y, z], the ideal I = (y2, yz, z2) is primary. Give R the grading with weights|y| = |z| = 1, and |x| = 0. If g /∈

√I = (y, z), then g has a degree zero term. If f /∈ I, then f

has a term of degree zero or one. The product has a term of degree zero or one, so is not in I.

3. In R = C[x, y, z], the ideal q = (x2, xy) is not primary, even though√q = (x) is prime. The

offending product is xy.

The definition of primary can be reinterpreted in many forms.

Proposition 4.29. If R is Noetherian, the following are equivalent:

1. q is primary.

2. Every zerodivisor in R/q is nilpotent.

3. Ass(R/q) is a singleton.

4. q has one minimal prime, and no embedded primes.

5.√q = p is prime and r ∈ R, w ∈ Rr p, rw ∈ q implies r ∈ q.

6.√q = p is prime, and qRp ∩R = q.

Proof. (1) ⇐⇒ (2): y is a zerodivisor mod q if there is some x /∈ q with xy ∈ q; the primaryassumption translates to a power of y is in q.

(2) ⇐⇒ (3): (2) translates to⋃

p∈Ass(R/q) p =⋂

p∈Min(R/q) p =⋂

p∈Ass(R/q) p. This holds if andonly if there is one associated prime.

(3) ⇐⇒ (4): is clear.(1) ⇐⇒ (5): Given the observation about that the radical of a primary ideal is prime, this is

just a rewording of the definition.(5) ⇐⇒ (6): We already know this from the discussion on behavior of ideals in localizations.

Remark 4.30. Let I be an ideal with√I = m a maximal ideal. If R is Noetherian, then AssR(R/I)

is nonempty and contained in Supp(R/I) = V (I) = {m}, so must be m, and hence I is primary.Note that the assumption that m is maximal was necessary here.


Next, we observe:

Lemma 4.31. If I1, . . . , It are ideals, then Ass(R/(⋂tj=1 Ij)) ⊆

⋃tj=i Ass(R/Ij). In particular, a

finite intersection of p-primary ideals is p-primary.

Proof. There is an injection R/(I1 ∩ I2) ↪→ R/I1 ⊕ R/I2. Hence, Ass(R/(I1 ∩ I2)) ⊆ Ass(R/I1) ∪Ass(R/I2); the statement for larger t is an obvious induction.

The latter statement follows from characterization #3 above.

Definition 4.32 (Primary decomposition). A primary decomposition of an ideal I is an expressionof the form

I = q1 ∩ · · · ∩ qt,

with each qi primary. A minimal primary decomposition of an ideal I is a primary decompositionas above in which

√qi 6=

√qj for i 6= j, and qi 6⊇

⋂j 6=i qj for all i.

Remark 4.33. By the previous lemma, we can turn a primary decomposition into a minimal oneby combining the terms with the same radical, then removing redundant terms.

Theorem 4.34 (Existence of primary decompositions). If R is Noetherian, then every ideal of Radmits a primary decomposition.

Proof. We will say that an ideal is irreducible if it cannot be written as a proper intersection of largerideals. If R is Noetherian, any ideal of R can be expressed as a finite intersection of irreducibleideals: if not, there would be an ideal maximal with the property of not being an intersectionof irreducible ideals, which must be an intersection of two larger ideals, each of which are finiteintersections of irreducibles, giving a contradiction.

Now, we claim that every irreducible ideal is primary. To establish the contrapositive, takexy ∈ q with x /∈ q, y /∈ √q. The ascending chain of ideals Jn = (q : yn) stabilizes for some n; thismeans that yn+1f ∈ q ⇒ ynf ∈ q. We will show that (q + (yn)) ∩ (q + (x)) = q.

The containment q ⊆ (q + (yn)) ∩ (q + (x)) is clear. If a is in the RHS, we have a = q + byn

for some q ∈ q, and ya ∈ q (since a ∈ q + (x)), so byn+1 ∈ q. Then, byn ∈ q by the hypothesison the chain Jn, and hence a ∈ q, as required. This shows that q is decomposable, concluding theproof.

There are also some uniqueness theorems for primary decomposition.

1 Octubre

Theorem 4.35 (First uniqueness theorem for primary decompositions). If I is an ideal in aNoetherian ring R, then for any minimal primary decomposition of I, I = q1 ∩ · · · ∩ qt, we have{√q1, . . . ,

√qt} = Ass(R/I). In particular, this set is the same for all minimal primary decompo-

sitions of I.

Proof. For any primary decomposition (minimal or not), we have Ass(R/I) ⊆⋃i Ass(R/qi) =

{√q1, . . . ,√qt} from the lemma on intersections above. We just need to show that in a minimal

decomposition as above, every pj :=√qj is an associated prime.

Let Ij =⋂i 6=j qi ⊇ I. Since the decomposition is minimal, the module Ij/I is nonzero, hence

has an associated prime a; let a be the annihilator of xj in Ij/I, with xj ∈ R. Since qjxj ⊆ I, wehave that qj is contained in the annihilator of xj , and since pj is the unique minimal prime of qj(and a is a prime containing qj), pj ⊆ a. On the other hand, if r ∈ a, we have rxj ∈ I ⊆ qj , andsince xj /∈ qj , we must have r ∈ pj by the definition of primary. Thus, a ⊆ pj , so a = pj .

4.3. PRIMARY DECOMPOSITION 49

We note that if we don’t assume that R is Noetherian, we may or may not have a primarydecomposition for a given ideal. It is true that if an ideal I in a general ring has a primarydecomposition, then the primes occurring are the same in any minimal decomposition. However,they are not the associated primes in general; rather, they are the primes that occur as radicals ofannihilators of elements.

There is also a partial uniqueness result for the actual primary ideals that occur in a minimaldecomposition.

Theorem 4.36 (Second uniqueness theorem for primary decompositions). If I is an ideal in aNoetherian ring R, then for any minimal primary decomposition of I, I = q1 ∩ · · · ∩ qt, the set ofminimal components {qi |

√qi ∈ Min(R/I)} is the same. Namely, qi = IR√qi ∩R.

Proof. We observe that a localization qa of a p-primary ideal q is either the unit ideal (if a 6⊆ p), oris a pa-primary ideal: this follows from the fact that the associated primes of R/q localize.

Now, since finite intersections commute with localization, for any prime a, we have

Ia = (q1)a ∩ · · · ∩ (qt)a

is a (not necessarily minimal) primary decomposition. In a minimal decomposition, choose aminimal prime a = pi to get Ipi = (qi)pi ; the other components in the intersection become theunit ideal since their radicals are not contained in pi. We can then contract to R to get Ipi ∩R =(qi)pi ∩R = qi, since qi is pi-primary.

Example 4.37. If R is Noetherian, and I is a radical ideal, we have that I = p1 ∩ · · · ∩ pt, wherepi are the minimal primes of I. This is the only primary decomposition of a radical ideal.

Example 4.38. Let R = K[x, y], where K is a field, and I = (x2, xy). We can write

I = (x) ∩ (x2, xy, y2) = (x) ∩ (x2, y).

These are two different minimal primary decompositions of I. To check this, we just need to seethat each of the ideals (x2, xy, y2) and (x2, y) are primary. Observe that each has radical m = (x, y),which is maximal, so by an earlier remark, these ideals are both primary.

Definition 4.39 (Symbolic power). If p is a prime ideal in a ring R, the nth symbolic power of pis p(n) := pnRp ∩R.

This admits equivalent characterizations.

Proposition 4.40. Let R be Noetherian, and p a prime ideal of R.

1. p(n) = {r ∈ R | ∃s /∈ p : rs ∈ pn}.

2. p(n) is the unique smallest p-primary ideal containing pn.

3. p(n) is the p-primary component in any minimal primary decomposition of pn.

Proof. The first characterization follows from the definition, and the fact that expanding andcontraction to/from a localization is equivalent to saturating with respect to the multiplicative set.

We know that p(n) is p-primary from a characterization of primary above. Any p-primary idealsatisfies qRp ∩R = q, and if q ⊇ pn, then p(n) = pnRp ∩R ⊆ qRp ∩R = q. Thus, p(n) is the uniquesmallest p-primary ideal containing pn.

The last characterization follows from the second uniqueness theorem above.


Example 4.41. 1. In R = K[x, y, z], p = (y, z), we have p(n) = pn for all n. This follows alongthe same lines as an example above.

2. In R = K[x, y, z] = (xy − z2), p = (y, z), we have p(2) 6= p2. Indeed, xy = z2 ∈ p2, and x /∈ p,so y ∈ p(2) r p2.

3. Let X = X3×3 be a 3 × 3 matrix of indeterminates, and K[X] be a polynomial ring over afield K. Let p = I2(X) be the ideal generated by 2 × 2 minors of X. We will write ∆i|k

j|lfor

the determinant of the submatrix with rows i, j and columns k, l. We find

x11 det(X) = x11x31∆1|22|3− x11x32∆1|1

2|3+ x11x33∆1|1

2|2

= (x11x31∆1|22|3− x11x32∆1|1

2|3+ x11x33∆1|1

2|2)− (x11x31∆1|2

2|3− x12x31∆1|1

2|3+ x13x31∆1|1

2|2)

= −∆1|13|2

∆1|12|3

+ ∆1|13|3

∆1|12|2∈ I2(X)2.

Note that in the second row, we subtracted the Laplace expansion of determinant of thematrix with row 3 replaced by another copy of row 1. That is, we subtracted zero.

Chapter 5

Spec and dimension

5.1 Dimension and height

Definition 5.1. • A chain of primes of length n in a ring R is

a0 $ a1 $ · · · $ an with ai ∈ Spec(R).

• A chain of primes as above is saturated if for each i, there is no q ∈ Spec(R) with ai ( q ( ai+1.

• The dimension or Krull dimension of a ring R is the supremum of the lengths of chains ofprimes in R. Equivalently, it is the supremum of the lengths of saturated chains of primesin R.

• The height of a prime p is the supremum of the lengths of chains of primes in R that end inp (i.e., p = an above). Equivalently, it is the supremum of the lengths of saturated chains ofprimes in R that end in p.

• The height of an ideal I is the infimum of the heights of the minimal primes of I.

To get a feel for these definitions, we make a sequence of easy observations.

Remark 5.2. 1. If p is prime, then dim(R/p) is the supremum of the lengths of (saturated)chains of primes in R

a0 $ a1 $ · · · $ an

with each ai ∈ V (p).

2. If I is an ideal, then dim(R/I) is the supremum of the lengths of (saturated) chains ofprimes in R

a0 $ a1 $ · · · $ an

with each ai ∈ V (I).

3. If W is a multiplicative set, then dim(W−1R) ≤ dim(R).

4. If p is prime, then height(p) = dim(Rp).

5. If q ⊇ p are primes, then dim(Rq/pRq) is the supremum of the lengths of (saturated) chainsof primes in R

a0 $ a1 $ · · · $ an

with a0 = p and an = q.

51

52 CHAPTER 5. SPEC AND DIMENSION

6. dim(R) = sup{height(m) | m ∈ Max(R)}.

7. dim(R) = sup{dim(R/p) | p ∈ Min(R)}.

8. If p is prime, dim(R/p) + height(p) ≤ dim(R).

9. If I is an ideal, dim(R/I) + height(I) ≤ dim(R).

10. A prime has height zero if and only if it is a minimal prime.

3 Octubre

We will need a few theorems before we can compute many examples, but we can handle a fewbasic cases.

Example 5.3. 1. The dimension of a field is zero.

2. A ring is zero-dimensional if and only if every minimal prime is maximal.

3. The ring of integers Z has dimension one, since there is one minimal prime (0) and everyother prime is maximal. Likewise, a principal ideal domain has dimension one.

4. If R is a UFD, I is a prime of height one if and only if I = (f) for a prime element f .

To see this, note that if I = (f) with f irreducible, and 0 $ p ⊆ I, then p contains somenonzero multiple of f , say afn with a and f coprime. Since a /∈ I, a /∈ p, so we must havef ∈ p, so p = (f). Thus, I has height one. On the other hand, if I is a prime of height one,we claim I contains an irreducible element. Indeed, I is nonzero, so contains some f 6= 0, andprimeness implies one of the prime factors of f is contained in I. Thus, any nonzero primecontains a prime ideal of the form (f), so a height one prime must be of this form.

5. It follows from the definition that if K is a field, then dim(K[x1, . . . , xd]) ≥ d, since there isa saturated chain of primes (0) $ (x1) $ (x1, x2) $ · · · $ (x1, . . . , xd).

We pose a related definition for modules.

Definition 5.4. The dimension of an R-module M is defined as dim(R/ annR(M)).

Note that if M is finitely generated, dim(M) is the same as the largest length of a chain ofprimes in SuppR(M).

Definition 5.5. • A ring is catenary if for every pair of primes q ⊇ p in R, every saturatedchain of primes

a0 $ a1 $ · · · $ an

with a0 = p and an = q has the same length.

• A ring is equidimensional if every maximal ideal has the same finite height, and everyminimal prime has the same dimension.

Example 5.6.K[x, y, z]

(xy, xz)is not equidimensional: note first that Min(R) = {(x), (y, z)}. We can see

this by computing Min((xy, xz)) in K[x, y, z]: (x) and (y, z) are prime, and (x)∩ (y, z) = (xy, xz),verifying the claim. Now, the height of (x−1, y, z) is one: it contains the minimal prime (y, z), andany saturated chain from (y, z) to (x− 1, y, z) corresponds to a saturated chain from (0) to (x− 1)in K[x], which must have length one since this is a PID. The height of (x, y − 1, z) is at least two,as witnessed by the chain (x) ⊆ (x, y − 1) ⊆ (x, y − 1, z).

5.2. OVER, UP, DOWN THEOREMS 53

Example 5.7. Z(2)[x] is a domain that is not equidimensional: consider the maximal ideal (2, x)that has height at least two, and the maximal ideal (2x − 1); this is maximal since the quotientis Q!

Remark 5.8. 1. If dim(R) <∞, R is a domain, and f 6= 0, then dim(R/(f)) < dim(R).

2. If R is equidimensional, then dim(R/(f)) < dim(R) if and only if f /∈⋃

p∈Min(R)

p.

3. In general, dim(R/(f)) < dim(R) if and only if f /∈⋃

p∈Min(R)dim(R/p)=dim(R)

p.

4. f /∈⋃

p∈Min(R) p if and only if dim(R/(p + (f))) < dim(R/p) for all p ∈ Min(R).

The one warning to make about dimension before we get too optimistic is that there are Noethe-rian rings of infinite dimension.

Example 5.9. Let R = K[x1, x2, . . . ]. R is clearly infinite-dimensional, but is Noetherian. Let

W = Rr ((x1) ∪ (x2, x3) ∪ (x4, x5, x6) · · · )

and S = W−1R. This ring has primes of arbitrarily large height, given by the images of thoseprimes we cut out from W . Thus, it has infinite dimension. The work is to show that this ring isNoetherian. We omit this argument here.

5.2 Over, up, down theorems

In this section, we will collect theorems of three forms about the spectrum of a ring: theorems thatassert that the map on Spec is surjective, and theorems about lifting chains of primes.

Recall that the fiber ring over p of a homomorphism ϕ : R→ S is given by

κϕ(p) = (Rr p)−1(S/pS),

and that

Spec(κϕ(p)) ∼= (ϕ∗)−1(p),

the subspace of Spec(S) consisting of primes that contract to p. As a special case, we write κ(p)for the fiber of the identity map; this is Rp/pRp, the residue field of the local ring Rp.

Lemma 5.10 (Image criterion). Let ϕ : R→ S be a ring homomorphism, and p ∈ Spec(R). Thenp ∈ im(ϕ∗) if and only if pS ∩R = p.

Proof. If pS ∩R = p, thenR

p=

R

pS ∩R↪→ S

pS,

so, localizing at (R r p), we get an injection κ(p) ↪→ κϕ(p). The latter ring is nonzero, so itsspectrum is nonempty. Thus, there is a prime mapping to p.

If pS ∩ R 6= p, then pS ∩ R % p (the other containment always holds). Then, if q ∩ R = p, wehave q ⊇ pS, so q ∩R % p.

Note that pS may not be prime, in general.


Example 5.11. Let R = C[xn] ⊆ S = C[x]. The ideal (xn − 1)R is prime, while (xn − 1)S =(∏n−1i=0 x−ζi)S, where ζ is a primitive nth root of unity, is not. However, each of its minimal primes

(x− ζi)S contracts to (xn − 1)R. Similarly, the ideal xnR is prime, while xnS is not radical.

Corollary 5.12. If R ⊆ S is a direct summand, then Spec(S)→ Spec(R) is surjective.

Proof. From the homework, we know that IS ∩R = I for all ideals in this case.

We want to extend the idea of the last corollary to work for all integral extensions. The keyidea is encapsulated in a definition.

Definition 5.13. Let R be a ring, S an R-algebra, and I an ideal.

• An element r of R is integral over I if it satisfies an equation of the form

rn + a1rn−1 + · · ·+ an−1r + an = 0 with ai ∈ Ii for all i.

• An element of S is integral over I if the same condition holds.

• The integral closure of I in R is I, the set of elements of R that are integral over I.

• Similarly, we write IS

for the integral closure of I in S.

We leave a little exercise for you.

Exercise 5.14. Let R ⊆ S, I be an ideal of S, and t be an indeterminate. Consider the ringsR[It] ⊆ R[t] ⊆ S[t].

1. IS

= {s ∈ S | st ∈ S[t] is integral over the ring R[It]}.

2. IS

is an ideal.

We note that in older texts and papers (e.g., Atiyah-Macdonald and Kunz) a different definitionis given for integral closure of an ideal. The one we use here is more-or-less universally accepted asthe correct notion.

Lemma 5.15 (Extension-contraction lemma for integral extensions). Let R ⊆ S be integral, and

I be an ideal of R. Then, IS ⊆ IS. Hence, IS ∩R ⊆ I.

Proof. Let x ∈ IS. We can write x =∑t

i=1 aisi with ai ∈ I. Taking S′ = R[s1, . . . , st], we alsohave x ∈ IS′. Thus, it suffices to show the statements in the case S is module-finite over R.

Let S =∑Rbi. We have xbi = (

∑k aksk)bi =

∑j aijbj with aij ∈ I. We can write this as a

matrix-acts-like-a-scalar equation xv = Av, where v = (b1, . . . , bu), and A = [aij ]. By the adjointtrick, we have det(xI − A)v = 0. Since we can assume b1 = 1, we have det(xI − A) = 0. The factthat this is the type of equation we want follows from the monomial expansion of the determinant:any monomial is a product of n terms where some of the are copies of x, and the rest are elementsof I.

The last statement follows from the fact that IS∩R = I, which is immediate from the definition.

Theorem 5.16 (Lying over). If R ⊆ S is an integral inclusion then Spec(S)→ Spec(R) is surjec-tive.

5.2. OVER, UP, DOWN THEOREMS 55

Proof. We observe that I ⊆√I. Thus, for p prime, by the previous lemma, pS ∩ R = p, and the

result follows from the image criterion.

Remark 5.17. Both “integral” and “inclusion” are important: the map R→ Rf is a nonintegralinclusion if f is a nonzerodivisor, and the image is the complement of V (f); the map R → R/(f)is an integral noninclusion, and the the image is V (f).

Theorem 5.18 (Incomparability). If ϕ : R→ S is integral, and q ⊆ q′ are such that ϕ∗(q) = ϕ∗(q′),then q = q′.

Proof. Since the map R→ R/ ker(R) is injective on spectra, we can replace R by the quotient andassume ϕ is an integral inclusion.

Now, if R ↪→ S is integral, then R/p ↪→ S/pS (injective by the lemma above) is integral; takean integral equation for a representative. Furthermore, localizing at (R r p) preserves integrality:if x ∈ S and w ∈ Rr p, then we have equations of the form

xn + r1xn−1 + · · ·+ rn = 0 =⇒ (

x

w)n +

r1

w(x

w)n−1 + · · ·+ rn

wn= 0.

We then have that κ(p) ↪→ κϕ(p) is integral. If q is a prime of κϕ(p), then κ(p) ⊆ κϕ(p)/q isan integral inclusion from a field to a domain, and by a lemma from a while ago, we must havethat κϕ(p)/q is a field. Therefore, κϕ(p) is zero-dimensional. Since there are no inclusions betweenprimes in κϕ(p), there are no inclusions between primes that contract to p.

8 Octubre

Corollary 5.19. If R→ S is integral, and S is Noetherian, then for any p ∈ Spec(R), only finitelymany primes contract to p. If, moverover, S =

∑ti=1Rsi is generated as a module by t elements,

then for any prime of R, at most t primes of S map to p.

Proof. For the first statement, this case the fiber ring κϕ(p) of any prime is also Noetherian, hencehas finitely many minimal primes. Every prime of the fiber is minimal, though.

for the second statement, we have κφ(p) =∑t

i=1 κ(p)si/1, and hence κφ(p) is a zero-dimensionalring that is a κ(p)-vector space of dimension at most t. We have a minimal primary decomposition(0)κφ(p) = q1 ∩ · · · ∩ qs where

√qi are distinct maximal ideals; these maximal ideals are in bijection

with the primes mapping to p in S. Thus, for i 6= j, V (qi + qj) = V (qi) ∩ V (qj) = ∅, soqi + qj = κφ(p), and by the Chinese Remainder Theorem, κφ(p) ∼= κφ(p)/q1 × · · · × κφ(p)/qs. Themap κ(p)→ κφ(p)/qi is nonzero for each i (since it is integral), so each factor is a κ(p)-vector space.Considering vector space dimension, find that the number of factors s is at most t.

Corollary 5.20. If R → S is integral, then height(q) ≤ height(q ∩ R) for any q ∈ Spec(S). Inparticular, dim(S) ≤ dim(R).

Proof. Given a chain of primes a0 $ · · · $ an = q in Spec(S), we can contract to R, and we get achain of distinct primes in Spec(R), so the height of the latter is at least as big.

Theorem 5.21 (Going up). If R → S is integral, then for every p $ p′ in Spec(R) and q inSpec(S) with q ∩R = p, there is some q′ ∈ Spec(S) with q $ q′ and q′ ∩R = p′.

Proof. Consider the map R/p→ S/q. This is integral, as we observed above. It is also injective, solying over applies. Thus, there is a prime a of S/q that contracts to the prime p′/p in Spec(R/p).We can write a = q′/q for some q′ ∈ Spec(S), and we must have that q′ contracts to p′.


Corollary 5.22. If R ⊆ S is integral, then dim(R) = dim(S).

Proof. We just need to show that dim(R) ≤ dim(S). Given a chain of primes p0 $ · · · $ pn inSpec(R), by lying over, there is a prime q0 ∈ Spec(S) contracting to p0. Then by going up, we haveq0 $ q1 with q1 ∩R = p1. Continuing, we can build a chain of distinct primes in S of length n.

Definition 5.23. A domain is normal if it is integrally closed in its field of fractions.

Lemma 5.24. A unique factorization domain is normal: in particular, a polynomial ring over afield is normal.

Proof. Let R be a UFD, and r/s ∈ frac(R) be integral over R. We can assume that r and s haveno common factor. Then we have for some ai’s in R

rn

sn+ a1

rn−1

sn−1+ · · ·+ an = 0 ⇒ rn = −(a1r

n−1s+ · · ·+ ansn).

Any irreducible factor of s must then divide rn, and hence divide r; if s is not a unit then thiscontradicts that there is no common factor. Thus, r/s ∈ R.

Lemma 5.25. Let R be a normal domain, x be an element integral over R in some larger domain.Let K be the fraction field of R, and f(t) ∈ K[t] be the minimal polynomial of x over K.

1. If x is integral over R, then f(t) ∈ R[t] ⊆ K[t].

2. If x is integral over a prime p, then f(t) has all of its nonleading coefficients in p.

Proof. Let x be integral over R. Fix an algebraic closure of K containing x, and let x1 =x, x2, . . . , xu be the roots of f . Since f(t) divides a monic equation for x, each xi is integralover R.

Let S = R[x1, . . . , xu] ⊆ K. This is a module-finite extension of R, so all of its elements areintegral over R. The coefficients of f(t) are elementary symmetric polynomials in the x’s, hencethey lie in S. But, S ∩K = R since R is normal. Thus, the first statement holds.

Now, let x be integral over p. All of the x’s are integral over p by the same argument asabove. Since each xj ∈ pS , any elementary symmetric polynomial in the x’s lies in pS . Thus, thenonleading coefficients lie in pS ∩R = p.

Theorem 5.26 (Going down). Suppose that R is a normal domain, S is a domain, and R ⊆ Sis integral. Then, for every p′ $ p in Spec(R) and q in Spec(S) with q ∩ R = p, there is someq′ ∈ Spec(S) with q′ $ q and q′ ∩R = p′. In a picture:

∃q′

��

⊆ q

��p′ ⊆ p

Proof. Let W = (Srq)(Rrp′) be the multiplicative set consisting of products of elements in Srqand Rr p′. Note that each of these sets contains 1, so each set is in the product. We want to showthat W ∩ p′S is empty. It will follow that W−1(S/p′S) = (S r q)−1κS(p′) has a prime ideal, andhence there is a prime of S contained in q contracting to p′.

To that end, suppose x ∈ p′S ∩W . Since x ∈ p′S, it is integral over p′, so write x = rs withr ∈ Rr p′, s ∈ S r q, and consider the minimal polynomial of x over frac(R):

h(x) = xn + a1xn−1 + · · ·+ an = 0.

5.3. NOETHER NORMALIZATION AND DIMENSION OF AFFINE RINGS 57

By the lemma above, each ai ∈ p′ ⊆ R. Then, since r ∈ K, substituting x = rs yields and dividingby rn yields a polynomial that, viewed as a polynomial in s, is irreducible. That is, the minimalpolynomial of s is

g(s) = sn +a1

rsn−1 + · · ·+ an

rn= 0.

Since s ∈ S, hence is integral over R, the lemma above says that each airi

=: vi ∈ R. Since r /∈ p′,and rivi = ai ∈ p′, we have vi ∈ p′, and the equation g(s) = 0 then shows that s ∈

√p′S. Since

q ∈ Spec(S) contains pS and hence p′S, we have s ∈√p′S ⊆ q. This is the desired contradiction.

Remark 5.27. We have used the fact that our rings are domains to put the theory of minimalpolynomials to use. The one hypothesis we can weaken is that the target is a domain: it sufficesto assume that it is torisonfree as a module over the source. Here’s why we can’t get rind of morehypotheses.

Let R = K[x] ⊆ S = K[x, y]/(xy, y2 − y). R is a normal domain, and the inclusion is integral:y2−y = 0 is an integral dependence relation for y over R, so S is generated by one integral element.Now, (1− y) is a minimal prime of S: y ∈ S r (1− y), so x goes to zero in the localization (sincexy = 0) and 1− y goes to zero in the localization (since y(1− y) = 0), so the localization is a copyof K, which has only one prime, (0). We have x = x− xy = x(1− y) ∈ (1− y), so the contractioncontains (x), so must be (x). But, by minimality, we can’t “go down” from (1− y) to a prime lyingover (0).

The normality hypothesis is important too. Take R = K[x(1−x), x2(1−x), y, xy] ⊆ S = K[x, y].The element x is integral over R: x(1− x) ∈ R is a recipe: x is a root of z2 − z − x(1− x). Notethat x is in the fraction field of R, so this element shows both that S is integral over R, and that Ris not normal. Now, q = (1−x, y) ⊆ S is a maximal ideal lying over the maximal ideal p generatedby the specified generating set in R. We have xS ∩ R = (x(1 − x), x2(1 − x), xy)R = p′, but weclaim that no prime contained in q lies over p′. Such a prime must contain x(1 − x) and xy, butnot x (this would make it the unit ideal), so must contain y and 1− x, and the contraction is thenp, which is too big!

Corollary 5.28. If R is a normal domain, S is a domain, and R ⊆ S is integral, then height(q) =height(q ∩R) for any q ∈ Spec(S).

Proof. We already know that ht(q) ≤ ht(q∩R). Now, take a maximal chain up to q∩R, and applygoing down to get a chain just as long that goes up to q.

10 Octubre

5.3 Noether normalization and dimension of affine rings

Lemma 5.29 (Making a pure-power leading term). 1. Let A be a domain, and f ∈ R = A[x1, . . . , xn]be a (not necessarily homogeneous) polynomial of degree at most N . The A-algebra automor-phism of R given by φ(xi) = xi + xN

n−in for i < n and φ(xn) = xn maps f to a polynomial

that, viewed as a polynomial in xn with coefficients in A[x1, . . . , xn−1], has leading term dxanfor some d ∈ A, a ∈ N.

2. Let K be an infinite field, and f ∈ R = K[x1, . . . , xn], with |xi| = 1 be a homogeneouspolynomial of degree N . There is a degree-preserving K-algebra automorphism of R given


by φ(xi) = xi + aixn for i < n and φ(xn) = xn that maps f to a polynomial that viewedas a polynomial in xn with coefficients in K[x1, . . . , xn−1], has leading term kxNn for some(nonzero) k ∈ K.

Proof. 1. The map φ sends a monomial term dxa11 · · ·xann to a polynomial with unique highest

degree term dxa1Nn−1+a2Nn−2+···+an−1N+ann . Since each ai is less than N in each monomial,

the map (a1, . . . , an) 7→ a1Nn−1 + a2N

n−2 + · · ·+ an−1N + an is injective when restricted tothe set of exponent tuples; thus, none of the terms can cancel. We find that the leading termis of the promised form.

2. We just need to show that the xN coefficient is nonzero for some choice of a’s. You cancheck that the coefficient of the xN term is f(−a1, . . . ,−an−1, 1). But if this, thought of as apolynomial in the a’s, is identically zero, then f must be the zero polynomial.

Theorem 5.30 (Noether Normalization). 1. Let A be a domain, and R be a finitely generatedA-algebra. Then, there is some nonzero a ∈ A and x1, . . . , xt ∈ R algebraically independentover A such that Ra is module-finite over Aa[x1, . . . , xt]. In particular, if A = K is a field,then R is module-finite over K[x1, . . . , xt].

2. Let K be an infinite field, and R be a finitely generated N-graded K-algebra with R0 = K andR = K[R1]. Then there are homogeneous elements x1, . . . , xt ∈ R1 algebraically independentover K such that R is module-finite over K[x1, . . . , xt].

Proof. We proceed by induction on the number of generators n of R over A, with the case n = 0trivial.

Now, suppose that we know the result for A-algebras generated by at most n − 1 elements. IfR = A[r1, . . . , rn], with r1, . . . , rn algebraically independent over A, we are done. Assume that thereis some relation on the r’s: there is some f(x1, . . . , xn) ∈ A[x1, . . . , xn] such that f(r1, . . . , rn) = 0.By taking an A-algebra automorphism (changing our generators), we can assume that f has leadingterm axNn (in terms of xn) for some a. Then, f is monic in xn after inverting a, so Ra is module-finite over Aa[r

′1, . . . , r

′n−1]. By hypothesis, Aab[r

′1, . . . , r

′n−1] is module-finite over Aab[r

′′1 , . . . , r

′′s ]

for some b ∈ A and r′′1 , . . . , r′′s that are algebraically independent over A. Since Rab is module-finite

over Aab[r′1, . . . , r

′n−1], we are done.

In the graded case, we use the graded part of the previous lemma.

Remark 5.31. There also exist Noether normalizations for quotients of power series rings overfields: after a change of coordinates, one can rewrite any nonzero power series in KJx1, . . . , xnK asa series of the form u(xdn + ad−1x

d−1n + · · · + a0) for a unit u and a0, . . . , ad−1 ∈ KJx1, . . . , xn−1K.

This is called Weierstrass preparation. The proof of the Noether normalization theorem proceedsin essentially the same way. Thus, given KJx1, . . . , xnK/I, we have some module-finite inclusion ofanother power series ring KJz1, . . . , zdK ⊆ KJx1, . . . , xnK/I.

Theorem 5.32. Let R be a finitely generated domain over a field K. Let K[z1, . . . , zd] be anyNoether normalization for R. Then, for any maximal ideal m of R, the length of any saturatedchain of primes from 0 to m is d. In particular, the dimension of R is d.

The same holds in a quotient of a power series ring over a field.

Proof. We prove by induction on d that for any finitely generated domain with a Noether nor-malization with d algebraically independent elements, any saturated chain of primes ending in amaximal ideal has length d.

5.3. NOETHER NORMALIZATION AND DIMENSION OF AFFINE RINGS 59

When d = 0, R is a domain that is integral over a field, hence is a field, so the statement followstrivially.

Pick a saturated chain0 $ q1 $ · · · $ qk = m

and consider the contractions to A = K[z1, . . . , zd]:

0 $ p1 $ · · · $ pk.

By the saturated condition, q1 has height 1, and so does p1, by going down. Since A is a UFD,p1 = (f) for some prime element f . After a change of variables, we can assume that f is monic inzd over K[z1, . . . , zd−1]. Then,

0 = q1/q1 $ q2/q1 $ · · · $ qk/q1 = m/q1

is a saturated chain in the affine domain R/q1 to the maximal ideal m/q1. Now, K[z1, . . . , zd−1] ⊆A/(f) ⊆ R/q1 are module-finite, and we can apply the induction hypothesis to say that the chainwe found in R/q1 has length d− 1, so k − 1 = d− 1, and k = d.

The same proof holds for quotients of power series rings.

Corollary 5.33. The dimension of the polynomial ring K[x1, . . . , xd] is d.

Corollary 5.34. If R is a K-algebra, the dimension of R is less than or equal to the minimal sizeof a generating set for R. If equality holds for some finite generating set, then R is isomorphic toa polynomial ring over K, and the generators are algebraically independent.

Proof. The first statement is trivial unless R is finitely generated, in which case we can writeR = K[f1, . . . , fs] ∼= K[x1, . . . , xs]/I for some ideal I. We have dim(R) ≤ s, for certain. If I 6= 0,then dim(R) < s, since the zero ideal is not contained in I.

Corollary 5.35. Let R be a finitely generated algebra oor a quotient of a power series ring over afield.

• R is catenary.

If additionally R is a domain, then

• R is equidimensional, and

• ht(I) = dim(R)− dim(R/I) for all ideals I.

Proof. Let p ⊆ q be primes in R. We can quotient out by p, and assume that R is a domain andp = 0. Fix a saturated chain C from q to a maximal ideal m. Given two saturated chains C ′, C ′′

from 0 to q, the concatenations C ′|C and C ′′|C are saturated chains from 0 to m, and hence musthave the same length. It follows that C ′ and C ′′ have the same length.

Equidimensionality is clear from the theorem.We have ht(I) = min{ht(p) | p ∈ Min(I)} and dim(R/I) = max{dim(R/p) | p ∈ Min(I)}. Thus,

it suffices to show the equality for prime ideals. Now, take a saturated chain of primes C from 0to p, and a saturated chain C ′ from p to a maximal ideal m. C has length ht(p) by catenarity anddefinition of height, C ′ has length dim(R/p) by the theorem, and C|C ′ has length dim(R) by thetheorem.

Corollary 5.36. If R is a finitely generated domain over a field K, then dim(R) = trdegK(frac(R)).


Proof. If R ⊆ S is module-finite, then frac(R) ⊆ frac(S) is algebraic, and hence they have the sametranscendence degree over K. In particular, if A = K[z1, . . . , zd] is a Noether normalization for R,

trdegK(frac(R)) = trdegK(frac(A)) = trdegK(K(z1, . . . , zd)) = d = dim(A) = dim(R).

15 Octubre

Example 5.37. We want to try out our dimension theorems to give a few different proofs that

R =K[x, y, z]

(y2 − xz)has dimension two, for a field K.

First: K[x, z] is a Noether normalization for R, so the dimension is 2.Second: we observe that y2 − xz is irreducible, e.g., by thinking of it as a polynomial in y and

applying Eisenstein’s criterion. Then (y2 − xz) is a prime of height one, so the dimension of R isdim(K[x, y, z])− height((y2 − xz)) = 3− 1 = 2.

Third: we want the use the characterization by transcendence dimension. To do this, wewill show that R ∼= K[u2, uv, v2] ⊆ K[u, v]. Let S = K[x, y, z], and consider the surjective K-algebra homomorphism φ : S → K[u2, uv, v2] given by φ(x) = u2, φ(y) = uv, φ(z) = v2. Clearly,(y2 − xz) ⊆ ker(φ); we claim this is an equality. To show this claim, first assume that K = K.We have ZK(ker(φ)) ⊆ ZK((y2 − xz)). Let (a, b, c) ∈ ZK((y2 − xz)). Let α, γ ∈ K be such thatα2 = a, γ2 = c; we can do this since K is algebraically closed. Then b2 = ac = (αγ)2. Replacingγ with −γ if necessary, we can assume that in fact b = αγ. Then (a, b, c) = (α2, αγ, γ2), soif f ∈ ker(φ), f(a, b, c) = f(α2, αγ, γ2) = φ(f)(α, γ) = 0. Thus, ZK(ker(φ)) = ZK((y2 − xz)), so√

ker(φ) =√

(y2 − xz). But ker(φ) and (y2−xz) are radical, so the claim holds in the algebraicallyclosed case. In general, we observe that S = K[x, y, z] ⊆ S = K[x, y, z] is faithfully flat. Then

S ⊗Skerφ

(y2 − xz)S∼=

(kerφ)⊗ S(y2 − xz)S ⊗ S

∼=ker (φ⊗ S)

(y2 − xz)S= 0

by the algebraically closed field case, sokerφ

(y2 − xz)S= 0, establishing the claim in general. Now,

frac(K[u2, uv, v2]) = K(u2, uv, v2) has u2, v2 as a transcendence basis over K, so the dimension is 2.

Chapter 6

Dimension, locally

6.1 Local rings and NAK

Definition 6.1. A ring R is called a local ring if it has exactly one maximal ideal. We often usethe notation (R,m) to denote R and its maximal ideal, or (R,m, k) to also specify the residue fieldk = R/m. Some people reserve the term local ring for a Noetherian local ring, and call what wehave defined a quasilocal ring; we will not follow this convention here.

An easy equivalent characterization is that R is local if and only if the set of nonunits of Rforms an ideal: this must then be the unique maximal ideal.

The following remark is an easy source of local rings.

Remark 6.2. If R is a ring and p is a prime ideal, then (Rp, pRp) is a local ring. Indeed, theprimes of Rp are just the expansions of primes of R that are contained in p. In R, p is uniquelymaximal among primes contained in p.

Example 6.3. 1. The ring Z/(pn) is local with maximal ideal (p).

2. The ring Z(p) = {ab ∈ Q | p - b when in lowest terms} is a local ring with maximal ideal (p).

3. The ring of power series KJxK over a field K is local. Indeed, a power series has an inverse ifand only if its constant term is nonzero. The complement of this set of units is an ideal (theideal (x)).

4. The ring of complex power series holomorphic at the origin, C{x} is local. In the above setting,one proves that the series inverse of a holomorphic function at the origin is convergent on aneighborhood of 0.

5. A polynomial ring over a field is certainly not local; you know so many maximal ideals! Alocal ring we will often encounter is K[x1, . . . , xd](x1,...,xd). We can consider this as the ring ofrational functions that in lowest terms have a denominator with nonzero constant term. (Wecan talk about lowest terms since the polynomial ring is a UFD.)

6. Extending the following example, we have local rings like (K[x1, . . . , xd]/I)(x1,...,xd). If K isalgebraically closed and I is a radical ideal, then K[x1, . . . , xd]/I = K[X] is the coordinatering of some affine variety, and (x1, . . . , xd) = m0 is the ideal defining the origin (as a pointin X ⊆ Kd). Then we call (K[x1, . . . , xd]/I)(x1,...,xd) = K[X]m0 the local ring of the point0 ∈ X; some people write OX,0. The radical ideals of this ring consist of radical ideals of

61

62 CHAPTER 6. DIMENSION, LOCALLY

K[X] that are contained in m0, which by the Nullstellensatz correspond to subvarieties of Xthat contain 0.

We want to make a quick observation about local rings: let (R,m, k) be local. Since k is aquotient of R, the characteristic of R must be a multiple of the characteristic of k; the kernel of themap from Z can only get bigger in the composition. Of course, with example 2 in mind, we mustthink of 0 as a multiple of any integer for this to make sense. Now k is a field, so its characteristicis 0 or p for a prime p. If char(k) = 0, then necessarily char(R) = 0. If char(k) = p, we claim thatchar(R) must be either 0 or a power of p. Indeed, if we write char(R) = pn ·m with m coprime top, note that p ∈ m, so if m ∈ m, we have 1 ∈ (p,m) ⊆ m, which is contradiction. Since R is local,this means that m is a unit. But then, pnm = 0 implies pn = 0, so the characteristic must be pn.We summarize and add one more observation.

Proposition 6.4. Let (R,m, k) be a local ring. Then one of the following holds:

1. char(R) = char(k) = 0. We say that R has equal characteristic zero.

2. char(R) = 0, char(k) = p for a prime p. We say that R has mixed characteristic (0, p).

3. char(R) = char(k) = p for a prime p. We say that R has equal characteristic p.

4. char(R) = pn, char(k) = p for a prime p and an integer n > 1.

If R is reduced, then one of the first three cases holds.

There are a range of statements the go under the banner of Nakayama’s Lemma a.k.a. NAK.

Proposition 6.5. Let R be a ring, I an ideal, and M a finitely generated R-module. If IM = M ,then

• there is an element r ∈ 1 + I such that rM = 0, and

• there is an element a ∈ I such that am = m for all m ∈M .

Proof. Let m1, . . . ,ms be a generating set for M . By assumption, we have equations

m1 = a11m1 + · · ·+ a1sms , . . . , ms = as1m1 + · · ·+ assms,

with aij ∈ I. Setting A = [aij ] and v = [xi] we have a matrix equations Av = v, and hence(id − A)v = 0. By the adjoint trick, we have det(id − A) kills each mi, and hence M . Sincedet(id−A) ≡ det(id) ≡ 1 mod I, this determinant is the element r we seek for the first statement.

For the latter statement, set a = 1 − r; this is in I and satisfies am = m − rm = m for allm ∈M .

Example 6.6. We will use this to give a quick proof of the fact that, if M is a finitely generatedR-module, and ϕ : M → M is a surjective R-linear endomorphism, then ϕ is an isomorphism. Inthis setting, M is an R[x]-module by the rule xm = ϕ(m), x2m = ϕ2(m), etc. The hypothesis thatϕ is surjective says that xM = M . Then, some element of (x) acts as the identity on M : f(x)x isthe identity, so f(ϕ) ◦ ϕ is the identity. Thus, ϕ is an isomorphism.

Proposition 6.7. Let (R,m, k) be a local ring, and M be a finitely generated module. If M = mM ,then M = 0.

6.1. LOCAL RINGS AND NAK 63

Proof. By the previous lemma, there exists an element r ∈ 1 + m that annihilates M . Such an rmust be a unit, so 1 annihilates M ; i.e., M = 0.

Proposition 6.8. Let (R,m, k) be a local ring, and M be a finitely generated module. Form1, . . . ,ms ∈M ,

m1, . . . ,ms generate M ⇐⇒ m1, . . . ,ms generate M/mM.

Thus, any generating set for M consists of at least dimk(M/mM) elements.

Proof. The implication ⇒ is clear. Let N = 〈m1, . . . ,ms〉 ⊆ M . We have that M/N = 0 iffM/N = m(M/N) iff M = mM +N iff M/mM is generated by the image of N .

Definition 6.9. Let (R,m) be a local ring, and M a finitely generated module. A set of elements{m1, . . . ,mt} is a minimal generating set of M if the images of m1, . . . ,mt form a basis for theR/m vector space M/mM .

Observe that any generating set for M contains a minimal generating set, and that everyminimal generating set has the same cardinality.

We now want to give graded analogues for the results above.

Proposition 6.10. Let R be an N-graded ring, and M a Z-graded module such that [M ]<a = 0 forsome a. If M = (R+)M , then M = 0.

In particular, if M is a finitely generated Z-graded module and M = (R+)M , then M = 0.

Proof. If M is finitely-generated, then it can be generated by finitely generated homogeneouselements (the homogeneous pieces of some finite generating set); thus the first statement impliesthe second.

M lives in degrees at least a, but (R+)M lives in degrees strictly bigger than a. If M has anonzero element, it has a nonzero homogeneous element, and we obtain a contradiction.

Just as above, we obtain the following:

Proposition 6.11. Let R be an N-graded ring, with R0 a field, and M a Z-graded module suchthat [M ]<a = 0 for some a. A set of elements of M generates M if and only if their images inM/(R+)M spans as a vector space. Since M and (R+)M are graded, M/(R+)M admits a basis ofhomogeneous elements.

In particular, if K is a field, R is a positively graded K-algebra, and I is a homogeneous ideal,then I has a minimal generating set by homogeneous elements, and this set is unique up to K-linearcombinations.

Note that we can use NAK to prove that certain modules are finitely generated in the gradedcase; in the local case, we cannot.

We want to prove one more important fact about Noetherian local rings. We prepare with alemma.

17 Octubre

Lemma 6.12 (Radical lemma for finitely generated ideals). If I ⊆ J are ideals, J ⊆√I and J is

finitely generated, then there is some n with Jn ⊆ I.Thus, if R is Noetherian, for every ideal I, there is some n with

√In ⊆ I.


Proof. Write J = (f1, . . . , fm). By definition, there are a1, . . . , am with faii ∈ I. Letn = a1 + · · ·+ am + 1; by the pigeonhole principle, any product of at most n fi’s must lie in I.

For the second statement, just use the fact that√I is finitely generated.

Theorem 6.13 (Krull intersection theorem). Let (R,m, k) be a Noetherian local ring. Then⋂n∈Nm

n = 0.

Proof. Let J =⋂n∈Nm

n. We will show that J ⊆ mJ , hence J = mJ , and thus J = 0 by NAK.Let mJ = q1∩· · ·∩qt be a primary decomposition. We claim that J ⊆ qi for each i. If

√qi 6= m,

pick x ∈ m r √qi. We have xJ ⊆ mJ ⊆ qi, with x /∈ √qi, so J ⊆ qi by definition of primary. Ifinstead

√qi = m, there is some N with mN ⊆ qi by the radical lemma for finitely generated ideals.

We then have J ⊆ mN ⊆ qi, and we are done.

6.2 Artinian rings

To prepare for our next big theorems in dimension theory, we need to understand the structureof zero-dimensional Noetherian rings. To get started on that, we will take a theorem on primarydecomposition for certain ideals in not necessarily Noetherian rings.

Theorem 6.14. Let R be a ring, not necessarily Noetherian. Let I be an ideal such that V (I) isa finite set of maximal ideals m1, . . . ,mt. Then, there is a primary decomposition I = q1 ∩ · · · ∩ qtand we also have I = q1 · · · qt, and R/I ∼= R/q1 × · · · ×R/qt.

Proof. First, we claim that IRmi is miRmi-primary. Indeed, note that (R/I)mi = Rmi/IRmi has aunique maximal ideal miRmi . Thus, if x, y ∈ Rmi are such that xy ∈ IRmi , and x /∈ miRmi , then xis a unit modulo Imi , so y ∈ IRmi . Then, the contraction of a primary ideal is primary (prove it!)so qi = IRmi ∩R is mi-primary, and I ⊆ q1 ∩ · · · ∩ qt. On the other hand, equality of these modulesis a local property; if p /∈ V (I), then both sides are the unit ideal in Rp, otherwise, in Rmi they areproper but equal. Thus, the primary decomposition.

The fact that this intersection is a product and the quotient ring is a direct product followsfrom the Chinese remainder theorem: V (qi + qj) = V (qi) ∩ V (qj) = ∅, so each pair of ideals iscomaximal.

Definition 6.15. A nonzero module is simple if it has no proper submodules. Equivalently, M issimple if it isomorphic to R/m for some maximal ideal m.

The nontrivial implication comes from the fact that any nonzero module contains a cyclicmodule, and if M ∼= R/I with I not maximal, we can surject to R/m for a maximal ideal containingI, which has a proper kernel. Note already that if R is local, any simple module is isomorphic tothe residue field.

Definition 6.16. A module M has finite length if it has a filtration of the form

0 = M0 ⊆M1 ⊆M2 ⊆ · · · ⊆Mn = M

with Mi+1/Mi simple for each i; such a filtration is called a composition series of length n. Wesay a omposition series is strict if Mi 6= Mi+1 for all i. Two composition series are equivalentif the collections of composition factors Mi+1/Mi are the same up to reordering. The length of afinite length module M , denoted `(M), is the minimum of the lengths of a composition series of M .

We recall the Jordan-Holder theorem and some of its consequences:

6.2. ARTINIAN RINGS 65

Theorem 6.17 (Jordan-Holder theorem). • For a module of finite length, any filtration can berefined to a composition series.

• For a module of finite length, any two composition series admit refinements that are equivalent.

• Every strict composition series for a fixed module of finite length is equivalent, and hence hasthe same length.

Some basic consequences of this theorem for length include:

• If M ⊆ N , then `(N) = `(M) + `(N/M). (Extend a composition series for M to one for N .)

• If 0 = M0 ⊆M1 ⊆M2 ⊆ · · · ⊆Mn = M , then `(M) =∑n−1

i=0 `(Mi+1/Mi).

• If M ⊆ N , then `(M) ≤ `(N), with equality only if M = N .

• If M has finite length, then any descending chain of submodules of M stabilizes. Likewise,any ascending chain of submodule stabilizes.

Remark 6.18. We also note that if M is annihilated by a maximal ideal m, so that M is anR/m-module, the length of M is equal to its dimension as an R/m-vector space. In particular,`(M/mM) = dimR/m(M/mM).

Example 6.19. Let R = R[x, y](x,y). Then M = R/m2 has length 3, since we have a compositionseries 0 ⊆ xM ⊆ (x, y)M ⊆M ; note that each is a copy of R/m. However, M is not an R/m-vectorspace.

Definition 6.20. A ring is Artinian if every descending chain of ideals eventually stabilizes. Amodule is Artinian if every descending chain of submodules eventually stabilizes.

Remark 6.21. 1. If R is an Artinian ring, then R/I is Artinian for any ideal I of R.

2. If R is an Artinian ring, then any nonempty family of ideals has a minimal element.

3. If M is an Artinian module, and N ⊆M , then N and M/N are Artinian.

Theorem 6.22. The following are equivalent:

1. R is Noetherian of dimension zero.

2. R is a finite product of local Noetherian rings of dimension zero.

3. R has finite length as an R-module.

4. R is Artinian.

Proof. (1)⇒(2): Since R is Noetherian of dimension zero, every prime is maximal and minimal,and there are thus finitely many. By the theorem from above, R decomposes as a direct productof Noetherian local rings, which all must have dimension zero.

(2)⇒(3): It suffices to deal with the case (R,m) is local. In this case, the maximal ideal is theunique minimal prime, so it consists of the nilpotents in R: m =

√(0). Since R is Noetherian, the

previous lemma yields mn = 0 for some n. If m = (f1, . . . , ft), with t finite since R is Noetherian,each mi/mi+1 is generated by {fa11 · · · f

att | a1 + · · · + at = i} as a (R/m)-vector space, hence has

finite length, so the total length of R is finite.(3)⇒(4): We have obsered this above.


(4)⇒(1): First we show that R has dimension zero. If p is any prime, then A = R/p is Artiniansince the ideals of R/p are in bijection with a subset of the ideals of R. Pick a ∈ A some nonzeroelement. The ideals

(a) ⊇ (a2) ⊇ (a3) ⊇ · · ·

stabilize, so an = an+1b for some b. Since A is a domain, ab = 1 in A, so a is a unit. Thus, R/p isa field, so every prime is maximal.

Second, note that there are only finitely many maximal ideals. Otherwise, consider the chain

m1 ⊇ m1 ∩m2 ⊇ m1 ∩m2 ∩m3 ⊇ · · · .

This stabilizes, so mn+1 ⊇ m1 ∩ · · · ∩mn ⊇ m1 · · ·mn. By distinctness, we can pick fi ∈ mi rmn+1,but then f1 · · · fn ∈ m1 · · ·mn rmn+1, which is a contradiction. Now, we apply the decompositiontheorem from earlier to conclude that R is a finite direct product of local rings of dimension zero.Since each of the factors is a quotient ring, each is Artinian. It suffices to show that each factor isNoetherian, so WLOG assume that (R,m) is local.

Now, to see R is Noetherian, m ⊇ m2 ⊇ m3 ⊇ · · · stabilizes again, so that mn = mn+1; we can’tapply NAK yet since we don’t know mn is finitely generated. If mn 6= 0, consider the family Sof ideals I ⊆ m such that Imn 6= 0; this contains m. Just as the Noetherian property guaranteesmaximal elements of nonempty families, the Artinian property guarantees minimal elements; takeJ minimal in S. For some x ∈ J , xmn 6= 0, and (x) ⊆ J ⊆ m, so J = (x) is principal by minimality.Now, xm(mn) = xmn+1 = xmn 6= 0, so xm ⊆ (x) is in the family S of ideals, and by minimality,(x) = m(x). NAK applies to this, so (x) = (0), contradicting that mn 6= 0. Then, we have

0 = mn ⊆ mn−1 ⊆ · · · ⊆ m ⊆ R,

and since the Artinian property descends to submodules and quotients, each factor has finite length.Thus, R has finite length, so ideals in R satisfy ACC, as required.

Example 6.23. Some Artinian local rings includeK[x, y]/(x2, y2), K[x, y]/(x2, xy, y2), and Z/(pn).

Example 6.24. Even though every Artinian ring is Noetherian and finite length, it is not truethat Artinian modules are always Noetherian or finite length. Let R = CJxK, and M = R[1/x]/R.Note that R[1/x] is the ring of Laurent series, so M is the module of “tails” of these functions. Itdoes not have finite length; it is not even finitely generated! Observe that any submodule N of Meither contains 1/xn for all n, or else there is a largest n for which 1/xn ∈ N , and N = R · 1/xnfor this n. The module R · 1/xn ⊆M has length n, so is Artinian, thus M is Artinian.

29 Octubre

Proposition 6.25. Let R be a Noetherian ring, and M be an R-module. M has finite length ifand only if it is finitely generated and all of its associated primes are maximal ideals of R.

Proof. If M has finite length, then it is Noetherian, hence finitely generated. Any compositionseries is a prime filtration, since the composition factors are cyclic quotients by maximal ideals,so all the associated primes of M must occur as factors in the composition series, hence everyassociated prime is maximal.

Conversely, if M is finitely generated, and every associated prime is maximal, then consider aprime filtration of M . Every factor in the prime filtration must contain an associated prime of M :

6.3. HEIGHT AND NUMBER OF GENERATORS 67

indeed, if q is a composition factor that does not contain M , then Mq = 0 since AssRq(Mq) = ∅ bybehavior of associated primes under localization, but we get a nonzero composition series of Mq bylocalizing the composition series of M , which is a contradiction. Hence every composition factormust correspond to a maximal ideal, so this is a composition series.

Definition 6.26. If (R,m, k) is local, a coefficient field for R is a subfield K ⊆ R such that themap K → R→ R/m ∼= k is an isomorphism.

Rings like K[x](x)/I have coefficient fields: the copy of K. Some rings without coefficient fieldsare Z(p), R[x](x2+1). Some rings have lots of them: C[x, y](x) contains C(y) and C(x + y), whichboth are coefficient fields!

Remark 6.27. If (R,m, k) is local with coefficient field K, then a finite length R-module M maynot be a k-module (it may not be killed by m), but it is a K-vector space by restriction of scalars,and `(M) = dimK(M).

6.3 Height and number of generators

Theorem 6.28 (Krull’s principal ideal theorem). Let R be a Noetherian ring, and f ∈ R. Then,every minimal prime of (f) has height at most one.

We note that this is stronger than the statement that the height of (f) is at most one: we recallthat that means that some minimal prime of (f) has height at most one.

Proof. If the theorem is false, so that there is some R, p, f with p minimal over (f) and ht(p) > 1,localize at p and mod out by a minimal prime to obtain a Noetherian local domain (R,m) ofdimension at least two in which m is the unique minimal prime of (f). In particular, R = R/(f) iszero-dimensional. Let q be a prime in between (0) and m.

Consider the symbolic powers q(n) of q. Our goal is to show that these stabilize in R. SinceR/(f) is Artinian, the descending chain of ideals

qR ⊇ q(2)R ⊇ q(3)R ⊇ · · ·

stabilizes. We then have, for some n and all m > n, that q(n)R ⊆ q(m)R. Pulling back to R,q(n) ⊆ q(m) + (f). For q ∈ q(n), write q = q′ + fr, with q′ ∈ q(m) ⊆ q(n), so that fr ∈ q(n).Since f /∈ q, r ∈ q(n). This yields q(n) = q(m) + fq(n). Thus, q(n)/q(m) = f(q(n)/q(m)), soq(n)/q(m) = m(q(n)/q(m)), and by NAK, q(n) = q(m) in R.

Now, if a ∈ q is nonzero, we have an ∈ qn ⊆ q(n) = q(m) for all m, so⋂m∈N q

(m) 6= 0. On the

other hand, q(m) ⊆ qmRq, and⋂m∈N q

(m) ⊆⋂m∈N q

mRq = 0 by Krull intersection. This is thecontradiction we seek.

We want to generalize this, but it is not so straightforward to run an induction. We will needa lemma that allows us to control the chains of primes we get.

Lemma 6.29. Let R be Noetherian, p $ q $ r be primes, and f ∈ r. Then there is some q′ withp $ q′ $ r and f ∈ q′.

Proof. We can quotient out by p and localize at r, and assume that r is the maximal ideal, and thatf is nonzero (for otherwise we are done); once we have succeeded in this case, we can pull back ourprime to R. Then, by the principal ideal theorem, minimal primes of (f) have height one, henceare not r; we can take q′ to be one of those.


Theorem 6.30 (Krull height theorem). Let R be a Noetherian ring, and I = (f1, . . . , fn) be anideal generated by n elements. Then every minimal prime of I has height at most n.

Proof. We proceed by induction on n. The case n = 1 is the principal ideal theorem.

Let I = (f1, . . . , fn) be an ideal, p be a minimal prime of I, and p0 ( p1 $ · · · $ ph = p be asaturated chain of length h ending at p. If f1 ∈ p1, then we can apply the induction hypothesis tothe ring R = R/(f1) and the ideal (f2, . . . , fn)R: the chain p1R $ · · · $ phR has length at mosth− 1 = n− 1, and we will be done. We use the previous lemma to replace our given chain with achain of the same length that satisfies this hypothesis.

In the given chain, let f1 ∈ pi+1 r pi. If whenever i > 0 we can decrease i, we can eventuallyget to the chain we want. To do this, we just need to apply the previous lemma with r = pi+1,q = pi, and p = pi−1.

Example 6.31. 1. The bound is certainly sharp: an ideal generated by n variables (x1, x2, . . . , xn)in a polynomial ring has height n. There are many other such ideals, like (u3 − xyz, x2 +2xz − 6y5, vx + 7vy) ∈ K[u, v, w, x, y, z]. An ideal of height n generated by n elements iscalled a complete intersection.

2. The ideal (xy, xz) in K[x, y, z] has minimal primes of heights 1 and 2.

3. It is possible to have associated primes of height greater than the number of generators. For acheap example, in R = K[x, y]/(x2, xy), the ideal generated by zero elements (the zero ideal)has an associated prime of height two, namely (x, y).

4. For the same phenomenon, but in a nice polynomial ring, I = (x3, y3, x2u + xyv + y2w) ⊂R = K[u, v, w, x, y]. Note that (u, v, w, x, y) = (I : x2y2), so I has an associated prime ofheight 5.

5. Noetherian is necessary. Let R = K[x, xy, xy2, . . . ] ⊆ K[x, y]. Note that (x) is not prime: fora > 0, xya /∈ (x), since ya /∈ R, but (xya)2 = x · xy2a ∈ (x). Thus, m = (x, xy, xy2, . . . ) ⊆√

(x), and since m is a maximal ideal, we have equality, so Min (x) = {m}. However, theideal p = (xy, xy2, xy3, . . . ) = (y)K[x, y] ∩R is prime, and the chain (0) ( p ( m shows thatht(m) > 1.

Lemma 6.32. Let R be a Noetherian ring, and I be an ideal. Let f1, . . . , ft ∈ I, and Ji = (f1, . . . , fi)for each i. If fi /∈

⋃a∈Min(Ji−1)rV (I)

a for each i, then any minimal prime of Ji either contains I, or

else has height i.

Proof. By induction on i. For i = 0, J0 = (0), and every minimal prime has height zero.

If we know the statement for i = m, consider a minimal prime q of Jm+1. Since Jm ⊆ Jm+1,q must contain a minimal prime of Jm, say p. If p ⊇ I, then q ⊇ I. If p does not contain I,it has height m by the induction hypothesis. Then, fm+1 ∈ Jm+1 implies fm+1 ∈ q, but sincep ∈ Min(Jm)r V (I), fm+1 /∈ p, so q % p. Thus, the height of q is strictly greater than m, and byKrull height, it is then exactly m+ 1.

31 Octubre

Theorem 6.33. Let R be a Noetherian ring of dimension d.

6.3. HEIGHT AND NUMBER OF GENERATORS 69

1. If p is a prime of height h, then there are h elements f1, . . . , fh ∈ p such that p is a minimalprime of (f1, . . . , fh).

2. If I is any ideal in R, then there are (at most) d + 1 elements f1, . . . , fd+1 ∈ I such thatI =

√(f1, . . . , fd+1).

3. If R is local or graded (N-graded, with R0 a field), and I is an ideal (homogeneous in thegraded case), then there are d elements (homogeneous in the graded case) f1, . . . , fd ∈ I suchthat I =

√(f1, . . . , fd).

Proof. We will use the notation from the previous lemma.

1. If p is a minimal prime, then we take the “empty sequence:” p is minimal over (0). Otherwise,we will use the recipe from the lemma above, with I = p. We need to show that we can chooseh elements satisfying the hypotheses, i.e., that p 6⊆

⋃Min(Ji)rV (p) a for i = 0, . . . , h− 1. Note

first that unless i ≥ h, Min(Ji) cannot meet V (p), by Krull height, so Min(Ji) r V (p) =Min(Ji) in the specified range. If p ⊆

⋃Min(Ji)

a, then by prime avoidance, p is contained in aminimal prime of Ji, which cannot happen for i < h. Thus, we can choose (f1, . . . , fh) as in thelemma, and its minimal primes have height h, or else contain p. Since Jh = (f1, . . . , fh) ⊆ p,some minimal prime q of Jh is contained in p. We know that this q either contains p, andhence is p, or else is contained in and has the same height as p, so again must be equal to p.

2. Again, we use the recipe from above. We again need to see that we can do this. Inductively,we are choosing elements inside of I, so each Ji is contained in I, and V (I) ⊆ V (Ji).

If for some i we have Min(Ji) r V (I) = ∅, then each minimal prime of Ji will lie in V (I),so V (Ji) ⊆ V (I), and equality holds, so the radicals are equal. If Min(Ji)r V (I) 6= ∅, thenI 6⊆ q for any q ∈ Min(Ji) r V (I), and I 6⊆

⋃Min(Ji)rV (I) q by prime avoidance, so we can

choose elements as in the lemma.

Thus, by the lemma, we either end up with√

(f1, . . . , fi) =√I for i ≤ d or we get elements

(f1, . . . , fd+1) = Jd+1 ⊆ I such that the minimal primes contain I or have height at leastd+ 1. In the latter case, by the assumption on the dimension, no prime has height d+ 1, soall the minimal primes of Jd+1 contain I. But, since Jd+1 ⊆ I, a minimal prime of Jd+1 mustalso be minimal over I. Thus, Min(Jd+1) ⊆ Min(I), so V (Jd+1) ⊆ V (I), and equality holds,so the radicals are equal.

3. We run the same argument as in the last part (using homogeneous prime avoidance in thegraded case). The point is that the only (homogeneous, in the graded case) ideal of height dalready contains I.

Corollary 6.34. Let (R,m, k) be a Noetherian local ring. Then,

dim(R) = min{n | ∃f1, . . . , fn :√

(f1, . . . , fn) = m} ≤ dimk(m/m2).

That is, the dimension of a Noetherian local ring is bounded by minimal the number of generatorsof its maximal ideal.

In particular, a Noetherian local ring has finite dimension.

Proof. The dimension of a local ring is the height of its maximal ideal. Thus, by Krull height, theminimum n in the middle is at least dim(R), and the previous theorem gives the other direction.The second inequality just follows from the fact that the right-hand quantity is the minimal numberof generators of the ideal m.


Compare the last inequality to the fact that, for an algebra over a field, the dimension isbounded by the number of generators as a K-algebra. We also want to compare this with thecharacterization of the dimension of a vector space as the least number of linear equations neededto cut out the origin.

Definition 6.35. A sequence of d elements x1, . . . , xd in a d-dimensional Noetherian local ring(R,m) is a system of parameters or SOP if

√(x1, . . . , xd) = m.

A sequence of d homogeneous elements x1, . . . , xd in a d-dimensional N-graded finitely generatedK-algebra R, with R0 = K, is a homogeneous system of parameters if

√(x1, . . . , xd) = R+.

We say that elements x1, . . . , xt are parameters if they are part of a system of parameters; thisis a property of the set, not just the elements.

By the previous corollary, every local (or graded) ring admits a system of parameters, and thesecan be useful in characterizing the dimension of a local Neotherian ring, or the height of a primein a Noetherian ring. To help characterize systems of parameters, we pose the following definition:

Definition 6.36. Let R be a Noetherian ring. A prime p of R is absolutely minimal if dim(R) =dim(R/p).

Observe that an absolutely minimal prime is minimal, since dim(R) ≥ dim(R/p) + height(p).

Theorem 6.37. Let (R,m) be a Noetherian local ring, and x1, . . . , xt ∈ m.

1. dim(R/(x1, . . . , xt)) ≥ dim(R)− t.

2. x1, . . . , xt are parameters if and only if dim(R/(x1, . . . , xt)) = dim(R)− t.

3. x1, . . . , xt are parameters if and only if x1 is not in any absolutely minimal prime of R andxi is not contained in any absolutely minimal prime of R/(x1, . . . , xi−1) for each i = 2, . . . , t.

Proof. 1. If dim(R/(x1, . . . , xt)) = s, then take a system of parameters y1, . . . , ys forR/(x1, . . . , xt),and pull back toR to get x1, . . . , xt, y

′1, . . . , y

′s inR such that the quotient ofRmodulo the ideal

generated by these elements has dimension zero. By Krull height, we get that t+s ≥ dim(R).

2. Let d = dim(R). Suppose first that dim(R/(x1, . . . , xt)) = d − t. Then, there is a SOPy1, . . . , yd−t forR/(x1, . . . , xt); lift back toR to get a sequence of d elements x1, . . . , xt, y1, . . . , yd−tthat generate an m-primary ideal. This is a SOP, so x1, . . . , xt are parameters.

On the other hand, if x1, . . . , xt, are parameters, extend to a SOP x1, . . . , xd. If I is the imageof (xt+1, . . . , xd) in R′ = R/(x1, . . . , xt), we have R′/I is zero-dimensional, hence has finitelength, so AssR′(R

′/I) = {m}, and I is m-primary in R′. Thus, dim(R′) is equal to the heightof I, which is then ≤ d − t by Krull height. That is, dim(R′) ≤ d − t, and using the firststatement, we have equality.

3. This follows from the previous statement and the observation that dim(S/(f)) � dim(S) ifand only if f is not in any absolutely minimal prime of S.

It is worth comparing the characterization in part three with our existence proof: we constructeda system of parameters by inductively avoiding all the minimal primes; a system a parameters is asequence where we inductively avoid all of the absolutely minimal primes.

5 Diciembre

6.4. THE DIMENSION INEQUALITY 71

6.4 The dimension inequality

Our goal here is to prove a general inequality that we can think of as a generalization of thecharacterization of dimension of finitely generated algebras in terms of transcendence degree. Beforewe state and prove this, we give a generalization of our calculation of dimension of polynomial rings.

Theorem 6.38. Let R be a Noetherian ring, S = R[x] a polynomial ring in one variable over R,q ∈ Spec(S) and p = q∩R. Then height(q) ∈ {height(p),height(p) + 1}, and height(q) = height(p)if and only if q = pS.

Proof. If the height of p is at least h, take a chain of primes p0 $ · · · $ ph = p and expand top0R[x] $ · · · $ phR[x] = pR[x] ⊆ q. This is a chain of primes in S, so the height of q is at least theheight of p, and equality holds only if q = pR[x].

Now, we can localize at Rrp: this will not affect the height of p or of q, since any prime containedin q has empty intersection with this set. Thus, we assume (R, p) is local. Let (f1, . . . , fd) be asystem of parameters for p. To show that height(q) ≤ height(p) + 1, we want to show that there issome y such that (f1, . . . , fd, y) is a system of parameters for R[x]q. Thus, it suffices to show thatR[x]q/(x1, . . . , xd) has dimension at most one. This ring is a localization of (R/(f1, . . . , fd))[x], soit suffices to show that this latter ring has dimenion at most one. The nilradical of this ring is theideal p[x], and quotienting out by this does not affect the dimension, so the dimension is the sameas that of R/p[x], which is one, since R/p is a field.

If q = pS, we have R[x]q/(f1, . . . , fd) ∼= (R/(f1, . . . , fd))[x]p[x], and p[x] consists of nilpotentsin this ring, so it is the unique minimal prime; this localization has dimension zero. Thus, in thiscase, we have a system of parameters of the same size, so the heights are equal.

Corollary 6.39. If R is Noetherian, then dim(R[x1, . . . , xn]) = dim(R) + n.

Theorem 6.40. Let R ⊆ S be an inclusion of domains, with R Noetherian. Let q ∈ Spec(S) andp = q ∩R ∈ Spec(R). Then

height(q) + trdeg(κ(q)/κ(p)) ≤ height(p) + trdeg(frac(S)/frac(R)).

Proof. Case 1: S = R[s] is generated by one element.

Case 1(a): s is algebraically independent over R: Then S is isomorphic to a polynomial ringin the variable x, and trdeg(frac(S)/frac(R)) = 1. Consider the fiber κφ(p) of the inclusion φ: we

compute κφ(p) = (R r p)−1(R[x]pR[x]

)∼= κ(p)[x], which is a one-dimensional domain. Thus, either

q is minimal in κφ(p), in which case q = pR[x], or else qκφ(p) is a maximal ideal and contains anonzero polynomial in κ(p)[x]; from the definitions we have

κφ(p)

qκφ(p)∼=

(Rr p)−1(S/pS)

q(Rr p)−1(S/pS)∼= (Rr p)−1(S/q),

and since this is a field (in the case q 6= pR[x]), it must agree with κ(q), so κ(q) is generated as

a field over κ(p) by one algebraic element x. If q = pR[x], then κ(q) =(R[x]pR[x]

)pR[x]

∼= κ(p)(x), so

trdeg(κ(q)/κ(p)) = 1, and in this case height(q) = height(p), so equality holds. Otherwise, we haveheight(q) = height(p) + 1 by the theorem above and trdeg(κ(q)/κ(p)) = 0, so equality holds again.


Case 1(b): s is algebraic over R: Then S ∼= R[s] ∼= R[x]/I for a nonzero prime I, andtrdeg(frac(S)/frac(R)) = 0. Since I ∩ R = 0 and I 6= 0, the height of I is one, using the previoustheorem. Let q′ be the primage of q in R[x], so q′ ∩R = p. By the previous case, we know that

height(q′) + trdeg(κ(q′)/κ(p)) = height(p) + 1.

Thenheight(q′) ≥ height(q) + height(I) = height(q) + 1,

and κ(q′) = κ(q), so

height(q) + trdeg(κ(q)/κ(p)) ≤ height(q′)− 1 + trdeg(κ(q′)/κ(p)) = height(p),

as required.

Case 2: S = R[s1, . . . , sn] is finitely generated: inductively, and using case 1, we can assumewe know the result for algebras with at most n − 1 or 1 generators, let T = R[s1] and r = q ∩ T .Since S = T [s2, . . . , sn], we have

height(q) + trdeg(κ(q)/κ(r)) = height(r) + trdeg(frac(S)/frac(T )), and

height(r) + trdeg(κ(r)/κ(p)) = height(p) + trdeg(frac(T )/frac(R)).

Since transencedence degree is additive for field extensions, this case follows.

Case 3: No finite generation assumption: if we localize at R r p, we do not affect heights orchange fracton fields or residue fields, so we can assume that (R, p) is local. Let h, t ∈ N be suchthat h ≤ height(q) and t ≤ trdeg(κ(q)/κ(p)) (both might be infinite). Take a chain of primes in S:q0 $ q1 $ · · · $ qh = q. Let y1, . . . , yh be such that yi ∈ qi r qi−1 and z1, . . . , zt ∈ S be such thatthe images in κ(q) are algebrically independent over κ(p). Let S′ = R[y1, . . . , yh, z1, . . . , zt] ⊆ Sand q′ = q ∩ S. Note that height(q) ≥ h, and trdeg(κ(q′)/κ(p)) ≥ t. By Case 2, we have

h+ t ≤ height(q′) + trdeg(κ(q′)/κ(p)) ≤ height(p) + trdeg(frac(S′)/frac(R)),

and since this holds for all h ≤ height(Q) and t ≤ trdeg(κ(q)/κ(p)), the claimed inequality holds.

Corollary 6.41. Let R ⊆ S be domains, with R Noetherian. Then dim(S) ≤ dim(R)+trdeg(frac(S)/frac(R)).

Proof. The height of a prime in S plus some nonnegative number is bounded above by the heightof some other prime in R plus the transcendence degree.

7 Diciembre

Chapter 7

Hilbert functions

7.1 Hilbert functions of graded rings

Definition 7.1. If K is a field, the Hilbert function of an N-graded K-algebra R is the functionHR : Z 7→ N ∪ ∞ with values HR(t) := dimK([R]t). Similarly, if M is a Z-graded R-module, wedefine the Hilbert function of M in the same way.

Example 7.2. Let K be a field, and R = K[x, y]/(x2, y2). This ring has a K-vectors space basisby monomials that are not multiples of x2 and y2, namely {1, x, y, xy}. We then find HR(0) = 1,HR(1) = 2, HR(2) = 1, and HR(t) = 0 for t > 2.

The key example of a Hilbert function is that of a polynomial ring.

Example 7.3. Let K be a field, and R = K[x1, . . . , xn] be a polynomial ring with the standardgrading: |xi| = 1 for each i. To compute the Hilbert function, we need to compute the size of aK-basis for HR(t) for each t. We have

[R]t =⊕

a1+···+an=t

Kxa11 · · ·xann .

We can find a bijection between these monomials and the set of strings that contain t stars andn−1 bars, where the monomial xa11 · · ·xann corresponds to the string with a1 stars, then a bar, thena2 stars, a bar, etc. Thus, the number of monomials is the number of ways to choose n − 1 barsfrom t+ n− 1 spots, i.e.,

HR(t) =

(t+ n− 1

n− 1

)for t ≥ 0.

We observe the binomial function here can be expressed as a polynomial in t for t ≥ 0; let

Pn(t) =(t+ n− 1)(t+ n− 2) · · · (t+ 1)

(n− 1)!∈ Q[t].

Observe that Pn(t) has −1, . . . ,−(n− 1) as roots. Then we have

HR(t) =

{Pn(t) if t > −n0 if t < 0.

Note that the two cases overlap for t = −(n− 1), . . . ,−1.

73

74 CHAPTER 7. HILBERT FUNCTIONS

While the Hilbert function was a polynomial for n ∈ N in this example, this is not always thecase.

Example 7.4. Let K be a field, and R = K[x0, . . . , xn] be a polynomial ring with the standardgrading. Let f be an element of degree d. We will compute the Hilbert function of R/fR. Notefirst that we can apply −⊗K L for a larger field L to obtain a ring with the same Hilbert functionwith an infinite ground field. Then, after applying a degree-preserving coordinate change, we canassume that f is monic in the variable x0. Then R/fR has R′ = K[x1, . . . , xn] as a homogeneousNoether normalization, and R/fR = R′ · 1 + R′ · x0 + · · · + R′ · xd−1

0 as R′-modules. In fact, thisset is a free R′-module basis; any R′-relation on these elements would yield an algebraic relationon {x0, . . . , xn} of x0-degree less than d, which cannot be a multiple of (f). Thus, we have

[R′/fR]t = [R′]t ⊕ [R′]t−1 · x0 ⊕ · · · ⊕ [R′]t−d+1 · xd−10

as K-vector spaces for each t. We then have

HR/fR(t) = HR′(t) +HR′(t− 1) + · · ·+HR′(t− d+ 1).

If d ≤ n, then for all i = 0, . . . , d − 1 and all t ≥ 0 we have t − i > −n, so by the polynomialring example, we have HR′(t− i) = Pn(t− i) for each such i and all t ≥ 0, so the Hilbert functionagrees with a polynomial for all t ≥ 0.

Now, we suppose that d > n, so d − n − 1 ≥ 0. Using that Pn(t) = HR′(t) for t ≥ −n andPn(−n) 6= 0 = HR′(−n),

HR/fR(d− n− 1) = HR′(d− n− 1) +HR′(d− n− 2) + · · ·+HR′(−n+ 1) +HR′(−n)

= Pn(d− n− 1) + Pn(d− n− 2) + · · ·+ Pn(−n+ 1) + 0

6= Pn(d− n− 1) + Pn(d− n− 2) + · · ·+ Pn(−n+ 1) + Pn(−n).

However, for all t > d− n− 1, we have that

HR/fR(t) = Pn(t) + Pn(t− 1) + · · ·+ Pn(t− d+ 1).

Thus, if HR/fR(t) agrees with an element of Q[t] for all t ∈ N, it must be the polynomial Pn(t) +Pn(t− 1) + · · ·+ Pn(t− d+ 1), but since these are not equal for t = d− n− 1, HR/fR(t) does notagree with any element of Q[t] for all t ∈ N.

We summarize this example.

Proposition 7.5. Let R = K[x0, . . . , xn]/(f) where f is homogeneous of degree d in the standardgrading; note that n = dim(R). Then,

1. R is a free module of rank d over a (standard graded) homogeneous Noether normalization.

2. If d ≤ n, then HR(t) is a polynomial for all t ∈ N.

3. If d > n, then HR(t) is not a polynomial for all t ∈ N, but is a polynomial for t ≥ d− n.

We will see that the Hilbert function is always eventually equal to a polynomial, as in the lastexample. We prepare with a lemma.

Lemma 7.6. Let R be a graded ring, and

0→ L→M → N → 0

be a (degree-preserving) short exact sequence of graded R-modules. Then HM = HL +HN .

7.1. HILBERT FUNCTIONS OF GRADED RINGS 75

Proof. In every degree t we get short exact sequences of vector spaces 0→ [L]t → [M ]t → [N ]t → 0,and the claim follows.

Recall that the dimension of a module M is the dimension of R/ annR(M).

Theorem 7.7. Let K be a field, and R be a finitely graded K-algebra such that R0 = K and R isgenerated by elements of degree one. Let M be a finitely generated graded R-module. Then thereis a polynomial PM (t) ∈ Q[t] and some n ∈ N such that HM (t) = PM (t) for t ≥ n. Moreover,deg(PM ) = dim(M) − 1 and (dim(M) − 1)! times the leading coefficient is a positive integer; ifdim(M) = 0, then PM = 0.

Proof. We show by induction on the dimension of M that this holds for all rings R satisfyingthe hypotheses. If the dimension of M is zero, we claim that AssR(M) = (R+). Indeed, underthese hypotheses, every homogeneous ideal is contained in R+, since it is the ideal generated by allhomogeneous elements. Since every associated prime is homogeneous, it must be contained in R+.If some other homogeneous prime p is associated to M , we have dim(R/p) ≥ 1, and R/p ↪→ Mcontradicts the hypothesis on dimension. Since M is finitely generated and its only associatedprime is maximal, M has finite length. Thus, it must be finite dimensional as a vector space, soonly finitely many graded pieces can be nonzero.

Now, suppose M has dimension n. Take a homogeneous prime filtration of M :

M = Mm %Mm−1 %Mm−2 % · · · %M1 %M0 = 0

with Mi/Mi−1∼= R/pi(di) for some homogeneous primes pi and integers di. Using the lemma

inductively on i, we get that HM (t) = HR/p1(d1)(t)+· · ·+HR/pm(dm)(t). Observe that HR/pi(di)(t) =HR/pi(t+ di) for each i. Since the associated primes of M are contained in V (annR(M)), we havedim(R/pi) ≤ dim(M), and there must be some pi for which equality occurs, since every associatedprime of M occurs, and the minimal primes of annR(M) are associated to M . If we can show thateach module of the form R/pi verifies the conclusion of the theorem, then we are done: all of theclaims of polynomiality, degree, and positivity of leading term pass to HM (t) by the equality above,as the shifting does not change degree or the leading term, dim(M) = max{dim(R/pi)}, and theleading term satisfies the hypotheses again.

If M = R/pi, then take a homogeneous Noether normalization A for this K-algebra M , andconsider a homogeneous prime filtration for M as an A-module. Every factor is either a shift ofA, or else has dimension less than a := dim(A) = dim(M), since A is a domain. Applying theinduction hypothesis and the formula HM (t) = HR/p1(d1)(t)+ · · ·+HR/pm(dm)(t) from above to thiscontext, we find that HM (t) is a sum of shifts of the polynomial Pa(t) from the example above, pluspolynomials of lower degree. Thus, the claim holds for M , and hence in general as our induction iscomplete.

12 Diciembre

Definition 7.8. The Hilbert polynomial of a graded module is the polynomial PM (t) that agreeswith HM (t) for t � 0. The multiplicity of a nonzero M is (dim(M) − 1)! times the leadingcoefficient of PM (t), denoted e(M), which is a positive integer.

Proposition 7.9. Let K be a field, and R be a finitely graded K-algebra such that R0 = K andR is generated by elements of degree one. Let 0 → L → M → N → 0 be a short exact sequenceof graded R-modules. Then PM (t) = PL(t) + PN (t). If dim(L) = dim(M) = dim(N), thene(M) = e(L) + e(N).


Proof. Both claims follow from the fact that Hilbert functions are additive on short exact sequences.

Example 7.10. If R is a polynomial ring, then e(R) = 1. If R = S/fS for a polynomial ring Sand a homogeneous element f of degree d, then e(R) = d; we can compute this using the previouslemma plus the Noether normalization.

We also want to consider the case when the ring is not necessarily generated in degree one. Thekey fact we will use is the following.

Proposition 7.11. Let K be a field, and R be a finitely-generated positively graded K-algebra withR0 = K. Then there is some d ∈ N such that the subring R(d) =

⊕i∈N[R]id is generated as a

K-algebra by [R]d.

Proof. Exercise.

The definition we need to generalize our results is the following.

Definition 7.12. A function f : Z → R is called a quasipolynomial if there is an integer b andpolynomials p0, . . . , pb−1 ∈ R[t] such that f(n) = pc(n) for c ≡ n mod b for each n ∈ Z.

Theorem 7.13. Let K be a field, and R be a finitely graded K-algebra such that R0 = K. Let Mbe a finitely generated graded R-module. Then there is a quasipolynomial where each coefficient hasrational coefficients PM (t) such that HM (t) = PM (t) for t� 0.

Proof. Let d be such that R(d) is generated by [R]d. Thus, we can think of R(d) as a standard gradedK-algebra, if we consider the elements of [R]id to have degree i. Since R is finitely generated asa K-algebra, it is certainly a finitely generated R(d)-algebra. Furthermore, any element x ∈ Rsatisfies a monic equation of the form td−xd ∈ R(d)[t], so R is integral and module-finite over R(d).Thus, M is a finitely generated R(d)-module. However, its grading over R is not consistent withthe grading of R(d). We can decompose M as an R(d)-module as

M = M0 ⊕M1 ⊕ · · · ⊕Md−1, where Mj =⊕i∈N

[M ]j+id.

Then, setting [Mj ]i = [M ]j+id, we obtain a grading on each Mj that is compatible with R(d). Notealso, that each Mj is a submodule of a finitely generated module over a Noetherian ring, so is alsofinitely generated. Then, each Mj admits a Hilbert polynomial. Taking each of these, we obtain aquasipolynomial that agrees with the Hilbert function for large values.

7.2 Associated graded rings and general Hilbert functions

Definition 7.14. Let (R,m) be a local ring. The Hilbert function of R is the function HR(t) =`R(mt/mt+1).

This function has similar properties to the Hilbert function for graded rings. To see this, wewill use the notion of the associated graded ring.

Definition 7.15. The associated graded ring of an ideal I in a ring R is the ring grI(R) :=⊕n∈N I

n/In+1, with n-th graded piece In/In+1, and multiplication (a + In+1)(b + Im+1) = ab +Im+n+1 for a ∈ In, b ∈ Im. If (R,m) is local, then gr(R) := grm(R).

7.2. ASSOCIATED GRADED RINGS AND GENERAL HILBERT FUNCTIONS 77

Note that the multiplication is well-defined. We observe also from the definitions that [grI(R)]0 =R/I, so if (R,m, k) is local then [gr(R)]0 = k, and grI(R) is generated as a [gr(R)]0-algebra by ele-ments of degree one. Again by the definitions, we have that for (R,m) local, HR(t) = Hgr(R)(t). Toget a completely satisfactory analogue of the Hilbert function theory in this setting, we would liketo undertand the dimension of the associated graded ring. To understand this, we use the followingrelated object.

Definition 7.16. Let R be a ring, and I an ideal. The Rees ring of I is the N-graded ringR[It] ⊆ R[t], and the extended Rees ring of I is the Z-graded ring R[It, t−1] ⊆ R[t, t−1]. In bothcases, the grading is given by setting |t| = 1, and |r| = 0 for all r ∈ R.

To understand these rings better, we write out the graded pieces. We have

R[It] = R⊕ It⊕ I2t2 ⊕ · · · and R[It, t−1] = · · · ⊕Rt−2 ⊕Rt−1 ⊕R⊕ It⊕ I2t2 ⊕ · · · .

Note that t /∈ R[It, t−1] since 1 /∈ I, so t−1 is not a unit, even though it looks like one.

Lemma 7.17. There are isomorphisms R[It, t−1]/(t−1) ∼= grI(R) and R[It, t−1]/(t−1 − 1) ∼= R.

Proof. For the first isomorphism, since t is homogeneous, we can use the graded structure. Wehave

t−1R[It, t−1] = · · · ⊕Rt−2 ⊕Rt−1 ⊕ I ⊕ I2t⊕ I3t2 ⊕ · · · ,

and matching the graded pieces, we find that the claimed isomorphism holds.For the second isomorphism, we consider the map R[It, t−1]→ R given by sending t 7→ 1. This

is surjective, and the kernel is the set of elements amtm + · · ·+ ant

n such that am + · · ·+ an = 0.We claim that this ideal is generated by (t−1 − 1). We proceed by induction on n −m. The casen−m = 0 corresponds to there being at most one nonzero term, in which case any element in thekernel is zero, and the claim trivially holds. In n−m = 1, we have an element of the form atn−1−atnfor some a, which is of the form (atn)(t−1 − 1). For the inductive step, if am + · · ·+ an = 0, writeamt

m + · · · + antn = (amt

m + · · · + (an−1 + an)tn−1) + (−antn−1 + antn). Observe that the set of

coefficients in R[It, t−1] of tn is a subset of the set of coefficients of tn−1 so both terms in parentheseslive in R[It, t−1], and they are clearly in the kernel of the evaluation map. The inductive hypothesisapplies to each term in parentheses, so we are done.

Lemma 7.18. If R is a Noetherian ring, and I an ideal, then the minimal primes of R[It, t−1] areexactly the primes of the form pR[t, t−1] ∩R[It, t−1] for p ∈ Min(R).

Proof. Write (0) = q1 ∩ · · · ∩ qt, a minimal primary decomposition of (0) in R, and pi =√qi.

It is easy to see that qiR[t] is primary with radical piR[t] (check it!). Then the same is true inR[t, t−1], by localizing. Contracting to R[It, t−1], we get primary ideals that intersect to (0); noneis contained in the intersection of the others, since this is the case after intersecting with R, andlikewise the radicals are distinct since they contract to different primes in R.

Thus setting q′i = qiR[t, t−1]∩R[It, t−1], q′1 ∩ · · · ∩ q′t is a minimal primary decomposition of (0)in R[It, t−1].

Theorem 7.19. Let (R,m) be a Noetherian local ring, and I ⊆ m an ideal. Then dim(R) =dim(R[It, t−1])− 1 = dim(grI(R)).

Proof. By the previous lemma, we can reduce to the case that R is a domain for the first equality.Observe that trdeg(frac(R[It, t−1])/frac(R)) = trdeg(frac(R)(t)/frac(R)) = 1, so by the dimensioninequality, dim(R[It, t−1]) ≤ dim(R) + 1.


Now, we claim that

Q = · · · ⊕Rt−2 ⊕Rt−1 ⊕m⊕ It⊕ I2t2 ⊕ · · · = (m, It, t−1)R[It, t−1]

is a maximal ideal of height dim(R) + 1 in R[It, t−1]. The quotient ring is R/m, so it is clearlymaximal. Given a chain p0 $ · · · $ ph = m, of length h = dim(R), let qi = piR[t, t−1] ∩ R[It, t−1].Since qi ∩R = pi[t, t

−1] ∩R = pi this is a proper chain of primes in R[It, t−1]. We have

qh = · · · ⊕mt−2 ⊕mt−1 ⊕m⊕ It⊕ I2t2 ⊕ · · · = (m, It)R[It, t−1] $ Q

so the height of Q is at least dim(R) + 1, and hence equal to dim(R) + 1 using the previous upperbound on the dimension.

For the last equality, since t−1 is a nonzerodivisor on R[It, t−1], we have dim(grI(R)) ≤dim(R[It, t−1])− 1. For the other inequality, let Q = Q/(t−1). Then

dim(grI(R)) ≥ dim(grI(R)Q) = dim(R[It, t−1]Q/(t−1))

≥ dim(R[It, t−1]Q)− 1 = height(Q)− 1 = dim(R).

We now can summarize the behavior of Hilbert functions of local rings.

Theorem 7.20. Let (R,m, k) be a local ring. Then there is a polynomial PR(t) ∈ Q[t] of degreeequal to dim(R) − 1 such that HR(t) = PR(t) for t � 0, and (dim(R) − 1)! times the leadingcoefficient is a positive integer, called the multiplicity of R.

Proof. This follows from the equalitiesHR(t) = Hgr(R)(t) and dim(R) = dim(gr(R)), and that gr(R)is an N-graded k-algebra with degreee zero piece equal to k and finitely generated by elements ofdegree one.

Index

(R,m), 61

(R,m, k), 61

A[f1, . . . , fd], 5

HR(t), 73

K[X], 24

M(t), 44

Mf , 38

Mp, 38

RG, 13

Rf , 37

Rp, 37

T -graded, 15

T -graded module, 16

V (I), 25

VMax(I), 25

W−1M , 37

W−1α, 38

ZR(F ), 19

AssR(M), 43

BilR(M,N ;L), 33

C{z}, 10

HomR(L,α), 30

HomR(M,N), 30

HomR(α,L), 30

HomR−alg, 20

Max(R), 25

Min(I), 41

Spec(R), 25

Supp(M), 42

α⊗M , 33

α∗, 30

α∗, 30

κ(p), 53

κφ(p), 40

C(R,R), 11

C∞(R,R), 11

Sd, 13

adj(B), 8gr(R), 76grI(R), 76| r |, 15I, 54

IS

, 54p-primary ideal, 47√I, 23∑γ∈ΓAγ, 6

Bij , 8

ϕR, 6

absolutely minimal prime, 70affine variety, 23algebra, 5algebra-finite, 5algebraically independent, 6associated graded ring, 76associated prime, 43associated primes of an ideal, 43

basis, 6bilinear, 33

chain of primes, 51classical adjoint, 8coefficient field, 67complete intersection, 68composition series, 64contravariant functor, 31coordinate ring, 24covariant functor, 31

degree, 15degree-preserving, 15dimension, 51dimension of a module, 52direct summand, 16

79

80 INDEX

embedded prime, 46equal characteristic p, 62equal characteristic zero, 62equation of integral dependence, 7equivalent composition series, 64exact sequence, 29extension of scalars, 34

faithfully flat, 36fiber ring, 40filtration, 44fine grading, 15finitely generated A-algebra, 5finitely presented, 30flat algebra, 34flat homomorphism, 34flat module, 34free module, 6

Gaussian integers, 7generates as an algebra, 5

height, 51Hilbert function, 73, 76Hilbert polynomial, 75homogeneous element, 15homogeneous ideal, 15homogeneous system of parameters, 70homomorphism of algebras, 20

integral closure of A in R, 9integral closure of an ideal, 54integral element, 7integral over A, 7integral over an ideal, 54invariant, 13

Jacobian, 6

Krull dimension, 51

left exact sequence, 30length of a chain of primes, 51linear action, 13linearly reductive group, 17local ring, 61local ring of a point, 61localization of a module, 37localization of a ring, 37

map on Spec, 26maximal spectrum, 25minimal generators, 63minimal prime, 41mixed characteristic (0, p), 62module of homomorphisms, 30module-finite, 6multiplicatively closed subset, 26

Noetherian module, 11Noetherian ring, 9nonzerodivisor, 37normal domain, 56

parameters, 70presentation, 30primary decomposition, 48primary ideal, 47prime avoidance, 46prime filtration, 44prime spectrum, 25Puiseux series, 10purely transcendantal, 21

quasilocal ring, 61quasipolynomial, 76

radical ideal, 23radical of an ideal, 23reduced, 23relations, 30relations in algebra, 6restriction of scalars, 6right exact sequence, 29

saturated chain of primes, 51shift, 44short exact sequence, 29simple module, 64solution set, 19SOP, 70splitting, 16standard grading, 15strict composition series, 64subvariety, 23support, 42symbolic power, 49system of parameters, 70

total ring of fractions, 37

INDEX 81

transcendence basis, 21

transcendence degree, 21

Veronese ring, 14

weights, 15

· contents 0 conventions 3 1 finiteness conditions 5 1.1 finitely generated algebras . . . . . ....

Documents